PiTheory JEE Physics Practice & Revision Pack · Class 12 · Chapter 3 · JEE Main

Current Electricity

Chapter-wise practice and revision for JEE Main: concise notes on current and drift, resistivity, resistor networks, cells and Kirchhoff's laws, bridges, the potentiometer, and power, concept figures, and 50 selected previous-year JEE Main questions covering the major concepts and question patterns, each with a step-by-step solution.

50 selected questions7 topics5 concept figuresFree sample · complete chapter

How to study this chapter

Look at the topic's figures, then read its notes (5 to 10 minutes). They give the major equations and methods the questions use.
Solve the topic's questions with a timer: 2 minutes for Easy, 3 to 4 for Medium and Hard.
Check the solution. Read the Point to note line even if you were right: it is the idea you will reuse.
Mark each question Got it or Revise. Come back to the Revise list after two days.

Where the marks come from

Main idea of each of the 256 previous-year JEE questions in the PiTheory bank

Combinations74

Resistivity41

Cells31

Power26

Bridges25

Potentiometer21

Current19

1Electric current, drift speed and current densityQ1–Q6
2Resistance, resistivity and temperatureQ7–Q15
3Resistor combinations and metersQ16–Q23
4Cells, internal resistance and Kirchhoff's lawsQ24–Q32
5Wheatstone bridge and meter bridgeQ33–Q38
6PotentiometerQ39–Q43
7Electrical power and heatingQ44–Q50
✓Last-day checklist

Topic 1 · Q1–Q6

Electric current, drift speed and current density

19 of the 256 JEE PYQs in the bank test this as their main idea

Look first

Fast random motion, slow drift.

Quick notes

$$I=\frac{dq}{dt},\qquad I=neAv_d,\qquad J=\frac IA=nev_d=\sigma E,\qquad v_d=\frac{eE\tau}{m}=\mu E$$

Electrons drift slowly (mm/s) opposite to the field, with random thermal motion on top. Charge passed $=\int I\,dt$ (area under the $I$–$t$ graph).

Resistivity from the free-electron model: $\rho=\dfrac{m}{ne^2\tau}$; mobility $\mu=\dfrac{e\tau}{m}=\dfrac{v_d}{E}$.
Same $V$, same length: $v_d$ does not depend on area. In a wire of varying area, $I$ is the same everywhere, so $v_d\propto\dfrac1A$.
Higher temperature: $\tau$ falls, so $v_d$ falls (at fixed $E$).
Current through part of a cross-section: $I=\int J\,dA$.

Trap. Units: mm$^2=10^{-6}$ m$^2$.

Practice: 6 questions

Q1JEE Main 2024 (1 Feb, Shift 1)Easy

The current in a conductor is expressed as $I = 3t^2 + 4t^3$, where $I$ is in ampere and $t$ is in second. The amount of electric charge that flows through a section of the conductor from $t = 1\ \text{s}$ to $t = 2\ \text{s}$ is ______ C.

Numerical answer type

Answer and solution

Answer: 22

Concept. Current is the rate of flow of charge, so the charge is the area under the $I$–$t$ graph: $q=\int I\,dt$.

Step 1. Integrate. $\displaystyle\int_1^2(3t^2+4t^3)\,dt=\left[t^3+t^4\right]_1^2=(8+16)-(1+1)$.

Answer. $22$ C.

Point to note. $q=\int I\,dt$.

Q2JEE Main 2023 (30 Jan, Shift 1)Medium

The charge flowing in a conductor changes with time as $Q(t)=\alpha t-\beta t^2+\gamma t^3$, where $\alpha$, $\beta$ and $\gamma$ are constants. Minimum value of current is

A$\alpha-\dfrac{3\beta^2}{\gamma}$
B$\alpha-\dfrac{\gamma^2}{3\beta}$
C$\beta-\dfrac{\alpha^2}{3\gamma}$
D$\alpha-\dfrac{\beta^2}{3\gamma}$

Answer and solution

Answer: Option D: $\alpha-\dfrac{\beta^2}{3\gamma}$

Concept. $I=\dfrac{dQ}{dt}$. The minimum of $I(t)$ is where $\dfrac{dI}{dt}=0$.

Step 1. Current. $I=\alpha-2\beta t+3\gamma t^2$.

Step 2. Minimum. $\dfrac{dI}{dt}=-2\beta+6\gamma t=0\Rightarrow t=\dfrac{\beta}{3\gamma}$.

Step 3. Value. $I_{\min}=\alpha-\dfrac{2\beta^2}{3\gamma}+\dfrac{\beta^2}{3\gamma}=\alpha-\dfrac{\beta^2}{3\gamma}$.

Answer. Option D.

Point to note. $dI/dt=0$.

Q3JEE Main 2023 (8 Apr, Shift 1)Easy

A current of $2$ A flows through a wire of cross-sectional area $25.0\ \text{mm}^2$. The number of free electrons in a cubic metre are $2.0\times10^{28}$. The drift velocity of the electrons is ______ $\times10^{-6}\ \text{m s}^{-1}$ (given, charge on electron $=1.6\times10^{-19}$ C).

Numerical answer type

Answer and solution

Answer: 25

Concept. $I=neAv_d$: current equals (number per volume)(charge)(area)(drift speed).

Step 1. Solve. $v_d=\dfrac{2}{2\times10^{28}\times1.6\times10^{-19}\times25\times10^{-6}}=\dfrac{2}{80}=2.5\times10^{-2}$ mm/s.

Answer. $25\times10^{-6}$ m s$^{-1}$, so $25$.

Point to note. $v_d=I/(neA)$.

Q4JEE Main 2023 (31 Jan, Shift 1)Easy

The drift velocity of electrons for a conductor connected in an electrical circuit is $V_d$. The conductor is now replaced by another conductor with same material and same length but double the area of cross-section. The applied voltage remains same. The new drift velocity of electrons will be

A$V_d$
B$\dfrac{V_d}{2}$
C$\dfrac{V_d}{4}$
D$2V_d$

Answer and solution

Answer: Option A: $V_d$

Concept. $v_d=\dfrac{eE\tau}{m}$ and $E=\dfrac VL$. The drift speed depends on the field, not on the area.

Step 1. Same $V$ and $L$. $E$ is unchanged, so $v_d$ is unchanged (the larger area simply carries more current).

Answer. $V_d$ (Option A).

Point to note. $v_d\propto E$, not area.

Q5JEE Main 2022 (27 Jun, Shift 1)Easy

The current density in a cylindrical wire of radius $4$ mm is $4\times10^6\ \text{A m}^{-2}$. The current through the outer portion of the wire between radial distances $\dfrac R2$ and $R$ is ______ $\pi$ A.

Numerical answer type

Answer and solution

Answer: 48

Concept. For uniform current density, current $=J\times$ area. The outer ring has area $\pi(R^2-R^2/4)$.

Step 1. Area. $\pi\times\tfrac34\times(4\times10^{-3})^2=12\pi\times10^{-6}$ m$^2$.

Step 2. Current. $4\times10^6\times12\pi\times10^{-6}=48\pi$ A.

Answer. $48$.

Point to note. $I=J\times$ ring area.

Q6JEE Main 2019 (10 Apr, Shift 1)Medium

A current of $5\ \text{A}$ passes through a copper conductor (resistivity $= 1.7\times10^{-8}\ \Omega\,\text{m}$) of radius of cross-section $5\ \text{mm}$. Find the mobility of the charges if their drift velocity is $1.1\times10^{-3}\ \text{m s}^{-1}$.

A$1.5\ \text{m}^2\text{V}^{-1}\text{s}^{-1}$
B$1.8\ \text{m}^2\text{V}^{-1}\text{s}^{-1}$
C$1.0\ \text{m}^2\text{V}^{-1}\text{s}^{-1}$
D$1.3\ \text{m}^2\text{V}^{-1}\text{s}^{-1}$

Answer and solution

Answer: Option C: $1.0\ \text{m}^2\text{V}^{-1}\text{s}^{-1}$

Concept. Mobility $\mu=\dfrac{v_d}{E}$; the field inside is $E=\rho J=\rho\dfrac{I}{A}$.

Step 1. Field. $E=\dfrac{1.7\times10^{-8}\times5}{\pi\times25\times10^{-6}}=1.08\times10^{-3}$ V/m.

Step 2. Mobility. $\mu=\dfrac{1.1\times10^{-3}}{1.08\times10^{-3}}\approx1.0$ m$^2$V$^{-1}$s$^{-1}$.

Answer. Option C.

Point to note. $\mu=v_d/(\rho J)$.

Topic 2 · Q7–Q15

Resistance, resistivity and temperature

41 of the 256 JEE PYQs in the bank test this as their main idea

Quick notes

$$R=\frac{\rho l}{A},\qquad R_t=R_0(1+\alpha\,\Delta t),\qquad \text{colour code: }R=(ab)\times10^{c}\ \pm\ \text{tolerance}$$

Colour digits: Black 0, Brown 1, Red 2, Orange 3, Yellow 4, Green 5, Blue 6, Violet 7, Grey 8, White 9. Tolerance: gold 5%, silver 10%, none 20%.

Change at constant volume	Resistance
Length $\times n$	$\times n^2$
Radius $\times\dfrac1n$	$\times n^4$
Small stretch $x\%$	$+2x\%$
Same mass, different radius	$R\propto1/r^4$

Hollow cylinder: $A=\pi(r_2^2-r_1^2)$. Radial flow between spheres: $R=\dfrac{\rho}{4\pi}\left(\dfrac1a-\dfrac1b\right)$; between cylinders: $\dfrac{\rho}{2\pi l}\ln\dfrac ba$.
Metals: $\rho$ rises with $T$. Semiconductors: falls. Alloys (manganin, constantan) have tiny $\alpha$: used in standard resistors.
Two equal resistors: series and parallel both have $\alpha=\dfrac{\alpha_1+\alpha_2}{2}$ (approximately).

Traps. Diameter vs radius. Resistivity depends only on material and temperature, not on shape.

Practice: 9 questions

Q7JEE Main 2024 (4 Apr, Shift 2)Medium

Two wires A and B are made of the same material and have the same mass. Wire A has a radius of $2.0\ \text{mm}$ and wire B has a radius of $4.0\ \text{mm}$. The resistance of wire B is $2\ \Omega$. The resistance of wire A is ______ $\Omega$.

Numerical answer type

Answer and solution

Answer: 32

Concept. Same material and same mass means the same volume, so $l\propto\dfrac1A$ and $R=\dfrac{\rho l}{A}\propto\dfrac{1}{A^2}\propto\dfrac{1}{r^4}$.

Step 1. Ratio. $\dfrac{R_A}{R_B}=\left(\dfrac{4}{2}\right)^4=16$.

Step 2. Value. $16\times2=32\ \Omega$.

Answer. $32$.

Point to note. Same mass: $R\propto1/r^4$.

Q8JEE Main 2024 (6 Apr, Shift 1)Easy

A wire of resistance $R$ and radius $r$ is stretched till its radius becomes $r/2$. If the new resistance of the stretched wire is $xR$, then the value of $x$ is ______.

Numerical answer type

Answer and solution

Answer: 16

Concept. Stretching keeps volume constant. Radius halves, so area becomes $\tfrac14$ and length $4$ times: $R\times16$.

Answer. $x=16$.

Point to note. $r/2$: $R\times16$.

Q9JEE Main 2023 (6 Apr, Shift 1)Easy

The length of a metallic wire is increased by $20\%$ and its area of cross-section is reduced by $4\%$. The percentage change in resistance of the metallic wire is ______.

Numerical answer type

Answer and solution

Answer: 25

Concept. $R=\dfrac{\rho l}{A}$. Use the given changes directly (the volume need not stay constant here).

Step 1. Ratio. $\dfrac{1.2}{0.96}=1.25$.

Answer. $25\%$ increase.

Point to note. $1.2/0.96$.

Q10JEE Main 2023 (12 Apr, Shift 1)Easy

A wire of resistance $160\ \Omega$ is melted and drawn in a wire of one-fourth of its length. The new resistance of the wire will be

A$16\ \Omega$
B$10\ \Omega$
C$640\ \Omega$
D$40\ \Omega$

Answer and solution

Answer: Option B: $10\ \Omega$

Concept. Melted and redrawn: volume constant. Length $\tfrac14$ means area $\times4$: $R\times\dfrac{1}{16}$.

Step 1. Value. $\dfrac{160}{16}=10\ \Omega$.

Answer. Option B.

Point to note. $l/4$: $R/16$.

Q11JEE Main 2023 (24 Jan, Shift 1)Easy

A hollow cylindrical conductor has length of $3.14\ \text{m}$, while its inner and outer diameters are $4\ \text{mm}$ and $8\ \text{mm}$ respectively. The resistance of the conductor is $n\times10^{-3}\ \Omega$. If the resistivity of the material is $2.4\times10^{-8}\ \Omega\,\text{m}$, the value of $n$ is ______.

Numerical answer type

Answer and solution

Answer: 2

Concept. For a hollow cylinder, use the area of the ring: $A=\pi(r_2^2-r_1^2)$, with radii from the diameters.

Step 1. Area. $\pi(16-4)\times10^{-6}=12\pi\times10^{-6}$ m$^2$.

Step 2. Resistance. $\dfrac{2.4\times10^{-8}\times3.14}{12\times3.14\times10^{-6}}=2\times10^{-3}\ \Omega$.

Answer. $n=2$.

Point to note. Ring area $\pi(r_2^2-r_1^2)$.

Q12JEE Main 2024 (4 Apr, Shift 1)Easy

The resistances of the platinum wire of a platinum resistance thermometer at the ice point and steam point are $8\ \Omega$ and $10\ \Omega$ respectively. After inserting it in a hot bath of temperature $400^\circ\text{C}$, the resistance of the platinum wire is

A$10\ \Omega$
B$8\ \Omega$
C$16\ \Omega$
D$2\ \Omega$

Answer and solution

Answer: Option C: $16\ \Omega$

Concept. $R_t=R_0(1+\alpha t)$. Two readings give $\alpha$; then predict.

Step 1. $\alpha$. $10=8(1+100\alpha)\Rightarrow\alpha=2.5\times10^{-3}\ ^\circ$C$^{-1}$.

Step 2. At $400^\circ$C. $8(1+1)=16\ \Omega$.

Answer. Option C.

Point to note. $\alpha$ from two readings.

Q13JEE Main 2024 (8 Apr, Shift 1)Easy

Resistance of a wire at $0\,{}^\circ\text{C}$, $100\,{}^\circ\text{C}$ and $t\,{}^\circ\text{C}$ is found to be $10\ \Omega$, $10.2\ \Omega$ and $10.95\ \Omega$ respectively. The temperature $t$ in Kelvin scale is ______.

Numerical answer type

Answer and solution

Answer: 748

Concept. Same law, $R_t=R_0(1+\alpha t)$; then convert to kelvin.

Step 1. $\alpha$. $\dfrac{0.2}{10\times100}=2\times10^{-4}$.

Step 2. Temperature. $10.95=10(1+2\times10^{-4}t)\Rightarrow t=475^\circ$C $=748$ K.

Answer. $748$.

Point to note. Then add 273.

Q14JEE Main 2019 (10 Apr, Shift 2)Medium

Space between two concentric conducting spheres of radii $a$ and $b$ $(b > a)$ is filled with a medium of resistivity $\rho$. The resistance between the two spheres will be:

A$\dfrac{\rho}{4\pi}\left(\dfrac1a + \dfrac1b\right)$
B$\dfrac{\rho}{2\pi}\left(\dfrac1a + \dfrac1b\right)$
C$\dfrac{\rho}{4\pi}\left(\dfrac1a - \dfrac1b\right)$
D$\dfrac{\rho}{2\pi}\left(\dfrac1a - \dfrac1b\right)$

Answer and solution

Answer: Option C: $\dfrac{\rho}{4\pi}\left(\dfrac1a - \dfrac1b\right)$

Concept. Current flows radially, so thin shells of area $4\pi r^2$ and thickness $dr$ are in series: $R=\int\dfrac{\rho\,dr}{4\pi r^2}$.

Step 1. Integrate. $\dfrac{\rho}{4\pi}\left(\dfrac1a-\dfrac1b\right)$.

Answer. Option C.

Point to note. Shells in series.

Q15JEE Main 2021 (18 Mar, Shift 2)Medium

Two wires of same length and thickness having specific resistances $6\ \Omega$ cm and $3\ \Omega$ cm respectively are connected in parallel. The effective resistivity is $\rho\ \Omega$ cm. The value of $\rho$, to the nearest integer, is ______.

Numerical answer type

Answer and solution

Answer: 4

Concept. Two equal wires in parallel are equivalent to one wire of the same length and double the area. Find the resistivity that gives the same resistance.

Step 1. Parallel value. $\dfrac{6\cdot3}{9}\dfrac LA=2\dfrac LA$.

Step 2. Equivalent wire. $\rho\dfrac{L}{2A}=2\dfrac LA\Rightarrow\rho=4\ \Omega$ cm.

Answer. $4$.

Point to note. Parallel = double area.

Topic 3 · Q16–Q23

Resistor combinations and meters

74 of the 256 JEE PYQs in the bank test this as their main idea

Quick notes

$$\text{series: }R=\sum R_i,\qquad \text{parallel: }\frac1R=\sum\frac1{R_i},\qquad \text{two in parallel: }\frac{R_1R_2}{R_1+R_2}$$

Series: same current, voltages in proportion to $R$. Parallel: same voltage, currents in inverse proportion to $R$.

Network	Result
Wire $R$ as regular $n$-gon, adjacent corners	$\dfrac{(n-1)R}{n^2}$
Wire $R$ as a circle, arc fraction $f$	$f(1-f)R$
$n$ equal resistors: max / min	$nR$ / $R/n$; ratio $n^2$
Wire cut into $n$ parts, all in parallel	$R/n^2$

Symmetry: points at the same potential can be joined or separated; the midpoint of an antisymmetric network is at $V/2$.
A real voltmeter is a resistor in parallel; it reads low. An ammeter is in series and should have tiny resistance.

Trap. Equivalent series resistance is larger than the largest; parallel is smaller than the smallest.

Practice: 8 questions

Q16JEE Main 2025 (24 Jan, Shift 1)Easy

A wire of resistance $9\ \Omega$ is bent to form an equilateral triangle. Then the equivalent resistance across any two vertices will be ______ ohm.

Numerical answer type

Answer and solution

Answer: 2

Concept. Between two vertices of a triangle, one side is in parallel with the other two sides in series.

Step 1. Sides. $3\ \Omega$ each.

Step 2. Combine. $\dfrac{3\times6}{9}=2\ \Omega$.

Answer. $2$.

Point to note. Side $\parallel$ two sides.

Q17JEE Main 2025 (28 Jan, Shift 1)Medium

A wire of resistance $R$ is bent into an equilateral triangle and an identical wire is bent into a square. The ratio of resistance between the two end points of an edge of the triangle to that of the square is

A$8/9$
B$27/32$
C$32/27$
D$9/8$

Answer and solution

Answer: Option C: $32/27$

Concept. Across one edge of a polygon: that edge in parallel with the rest. For $n$ sides, $R_{\text{edge}}=\dfrac{(n-1)R}{n^2}$.

Step 1. Triangle. $\dfrac{2R}{9}$. Square. $\dfrac{3R}{16}$.

Step 2. Ratio. $\dfrac{2/9}{3/16}=\dfrac{32}{27}$.

Answer. Option C.

Point to note. $(n-1)R/n^2$.

Q18JEE Main 2023 (1 Feb, Shift 2)Medium

Equivalent resistance between the adjacent corners of a regular $n$-sided polygon of uniform wire of resistance $R$ would be

A$\dfrac{(n-1)R}{n^2}$
B$\dfrac{(n-1)R}{2n-1}$
C$\dfrac{n^2R}{n-1}$
D$\dfrac{(n-1)R}{n}$

Answer and solution

Answer: Option A: $\dfrac{(n-1)R}{n^2}$

Concept. Between adjacent corners: $\dfrac Rn$ in parallel with $\dfrac{(n-1)R}{n}$.

Step 1. Combine. $\dfrac{(R/n)(n-1)R/n}{R}=\dfrac{(n-1)R}{n^2}$.

Answer. Option A.

Point to note. $(n-1)R/n^2$.

Q19JEE Main 2019 (9 Apr, Shift 2)Easy

A metal wire of resistance $3\ \Omega$ is elongated to make a uniform wire of double its previous length. This new wire is now bent and the ends joined to make a circle. If two points on this circle make an angle $60^\circ$ at the center, the equivalent resistance between these two points will be

A$\dfrac53\ \Omega$
B$\dfrac{12}{5}\ \Omega$
C$\dfrac72\ \Omega$
D$\dfrac52\ \Omega$

Answer and solution

Answer: Option A: $\dfrac53\ \Omega$

Concept. Doubling length (constant volume) gives $R\times4=12\ \Omega$. A $60^\circ$ arc is $\tfrac16$ of the circle; the two arcs are in parallel.

Step 1. Arcs. $2\ \Omega$ and $10\ \Omega$.

Step 2. Parallel. $\dfrac{20}{12}=\dfrac53\ \Omega$.

Answer. Option A.

Point to note. Stretch ×4, then arcs.

Q20JEE Main 2026 (24 Jan, Shift 2)Medium

A regular hexagon is formed by six wires each of resistance $r\Omega$ and the corners are joined to the centre by wires of same resistance. If the current enters at one corner and leaves at the opposite corner, the equivalent resistance of the hexagon between the two opposite corners will be

A$\dfrac35r$
B$\dfrac45r$
C$\dfrac58r$
D$\dfrac34r$

Answer and solution

Answer: Option B: $\dfrac45r$

Concept. Use symmetry. With $V$ at the entry corner A and $0$ at the opposite corner D, the centre O is at $V/2$; neighbours B, F of A are equipotential at $x$, and C, E are at $V-x$.

Step 1. Node B. Currents out of B sum to zero: $(x-V)+(x-\tfrac V2)+(x-(V-x))=0\Rightarrow x=\tfrac58V$.

Step 2. Current from A. To B and F: $2\cdot\dfrac{3V/8}{r}$; to O: $\dfrac{V/2}{r}$; total $\dfrac{5V}{4r}$.

Step 3. Resistance. $\dfrac{V}{I}=\dfrac45r$.

Answer. Option B.

Point to note. Symmetry: centre at $V/2$.

Q21JEE Main 2022 (24 Jun, Shift 2)Easy

What will be the most suitable combination of three resistors $A = 2\ \Omega$, $B = 4\ \Omega$, $C = 6\ \Omega$ so that $\left(\dfrac{22}{3}\right)\Omega$ is the equivalent resistance of the combination?

AParallel combination of $A$ and $C$ connected in series with $B$
BParallel combination of $A$ and $B$ connected in series with $C$
CSeries combination of $A$ and $C$ connected in parallel with $B$
DSeries combination of $B$ and $C$ connected in parallel with $A$

Answer and solution

Answer: Option B: Parallel combination of $A$ and $B$ connected in series with $C$

Concept. Test the options: $2\parallel4=\dfrac43$, then add $6$ in series.

Step 1. Value. $\dfrac43+6=\dfrac{22}{3}\ \Omega$.

Answer. Option B.

Point to note. Test options.

Q22JEE Main 2026 (24 Jan, Shift 1)Easy

Two resistors of $100\Omega$ each are connected in series with a 9 V battery. A voltmeter of $400\Omega$ resistance is connected to measure the voltage drop across one of the resistors. The voltmeter reading is ______ V.

A3
B2
C4
D4.5

Answer and solution

Answer: Option C: 4

Concept. A real voltmeter is a resistor in parallel with what it measures, so it lowers the reading.

Step 1. Parallel. $100\parallel400=80\ \Omega$.

Step 2. Divider. $9\times\dfrac{80}{180}=4$ V.

Answer. Option C.

Point to note. Voltmeter in parallel.

Q23JEE Main 2024 (30 Jan, Shift 2)Medium

Two resistances of $100\ \Omega$ and $200\ \Omega$ are connected in series with a battery of $4\ \text{V}$ and negligible internal resistance. A voltmeter is used to measure the voltage across the $100\ \Omega$ resistance, and it gives a reading of $1\ \text{V}$. The resistance of the voltmeter must be ______ $\Omega$.

Numerical answer type

Answer and solution

Answer: 200

Concept. Series circuit: voltages divide in proportion to resistance. Reading $1$ V means the other $3$ V is across $200\ \Omega$.

Step 1. Current. $\dfrac{3}{200}=0.015$ A.

Step 2. Parallel resistance. $\dfrac{1}{0.015}=\dfrac{200}{3}\ \Omega$.

Step 3. Voltmeter. $\dfrac1{R_V}=\dfrac{3}{200}-\dfrac1{100}=\dfrac1{200}$.

Answer. $200\ \Omega$.

Point to note. Find $R_V$ from the divider.

Topic 4 · Q24–Q32

Cells, internal resistance and Kirchhoff's laws

31 of the 256 JEE PYQs in the bank test this as their main idea

Look first

Part of the emf is lost inside the cell.

Quick notes

$$I=\frac{E}{R+r},\qquad V=E-Ir\ (\text{discharging}),\qquad V=E+Ir\ (\text{charging}),\qquad P_R\ \text{maximum at }R=r$$

Kirchhoff: currents into a junction sum to zero (charge conservation); emfs and drops around any loop sum to zero (energy conservation).

Combination	$E_{\text{eq}}$	$r_{\text{eq}}$
$n$ identical in series	$nE$	$nr$
$m$ identical in parallel	$E$	$r/m$
Two different in parallel	$\dfrac{E_1/r_1+E_2/r_2}{1/r_1+1/r_2}$	$\dfrac{r_1r_2}{r_1+r_2}$

Two identical cells give the same current in series and parallel when $R=r$.
Terminal voltage of one cell zero in series: $R=r_1-r_2$.
Two loads with the same cell: $E=I_1(R_1+r)=I_2(R_2+r)$.
In steady state a capacitor carries no current; it only reads the voltage across its ends.
Energy from a battery: $W=qV$; capacity in mAh $\times$ voltage gives the stored energy.

Traps. Reversed cell: change the sign of its emf only. Parallel cells: the equivalent emf lies between the two.

Practice: 9 questions

Q24JEE Main 2026 (5 Apr, Shift 2)Easy

When an external resistance of $5\ \Omega$ is connected across terminals of a cell, a current of $0.25$ A flows through it. When the $5\ \Omega$ resistor is replaced by a $2\ \Omega$ resistor, a current of $0.5$ A flows through it. The internal resistance of the cell is ______ $\Omega$.

Numerical answer type

Answer and solution

Answer: 1

Concept. $E=I(R+r)$ with the same $E$ and $r$ for both loads.

Step 1. Equate. $0.25(5+r)=0.5(2+r)\Rightarrow1.25+0.25r=1+0.5r$.

Answer. $r=1\ \Omega$.

Point to note. $E=I(R+r)$ twice.

Q25JEE Main 2026 (28 Jan, Shift 1)Medium

For the two cells having same EMF $E$ and internal resistance $r$, the current passing through the external resistor $6\Omega$ is same when both the cells are connected either in parallel or in series. The value of internal resistance $r$ is ______ $\Omega$.

Answer and solution

Answer: Option B: 6

Concept. Two identical cells: series gives $\dfrac{2E}{R+2r}$, parallel gives $\dfrac{E}{R+r/2}$.

Step 1. Equate. $2\left(6+\dfrac r2\right)=6+2r\Rightarrow r=6\ \Omega$.

Answer. Option B. (In general they are equal when $r=R$.)

Point to note. Equal when $r=R$.

Q26JEE Main 2024 (30 Jan, Shift 1)Medium

Two cells are connected in opposition in a single closed loop. Cell $E_1$ has $8\ \text{V}$ emf and $2\ \Omega$ internal resistance; cell $E_2$ has $2\ \text{V}$ emf and $4\ \Omega$ internal resistance. There is no other resistance in the circuit. The terminal potential difference of cell $E_2$ is ______ V.

Numerical answer type

Answer and solution

Answer: 6

Concept. Cells in opposition: net emf is the difference. The weaker cell is charged, so its terminal voltage is $E+Ir$.

Step 1. Current. $\dfrac{8-2}{2+4}=1$ A.

Step 2. Terminal voltage of $E_2$. $2+1\times4=6$ V.

Answer. $6$.

Point to note. Charging: $V=E+Ir$.

Q27JEE Main 2026 (6 Apr, Shift 2)Hard

Two cells of emfs $1$ V and $2$ V and internal resistance $2\ \Omega$ and $1\ \Omega$, respectively connected in parallel, gave a current of $1$ A through an external resistance. If the polarity of one cell is reversed, then value of current through the external resistance will be $\dfrac{\alpha}{5}$ A. The value of $\alpha$ is ______.

Numerical answer type

Answer and solution

Answer: 3

Concept. Cells in parallel act as one cell: $E_{\text{eq}}=\dfrac{E_1/r_1+E_2/r_2}{1/r_1+1/r_2}$, $r_{\text{eq}}=\dfrac{r_1r_2}{r_1+r_2}$. Reversing a cell changes the sign of its emf.

Step 1. Original. $E_{\text{eq}}=\dfrac{0.5+2}{1.5}=\dfrac53$ V, $r_{\text{eq}}=\dfrac23\ \Omega$; $1=\dfrac{5/3}{2/3+R}\Rightarrow R=1\ \Omega$.

Step 2. One reversed. $E_{\text{eq}}=\dfrac{-0.5+2}{1.5}=1$ V.

Step 3. Current. $\dfrac{1}{2/3+1}=\dfrac35$ A.

Answer. $\alpha=3$.

Point to note. Parallel cells: weighted emf.

Q28JEE Main 2022 (27 Jul, Shift 1)Medium

Two sources of equal emfs are connected in series. This combination is connected to an external resistance $R$. The internal resistances of the two sources are $r_1$ and $r_2$ $(r_1 > r_2)$. If the potential difference across the source of internal resistance $r_1$ is zero then the value of $R$ will be

A$r_1 - r_2$
B$\dfrac{r_1r_2}{r_1 + r_2}$
C$\dfrac{r_1 + r_2}{2}$
D$r_2 - r_1$

Answer and solution

Answer: Option A: $r_1 - r_2$

Concept. Terminal voltage of a cell $=E-Ir$. Zero means $Ir_1=E$.

Step 1. Current. $I=\dfrac{2E}{R+r_1+r_2}$.

Step 2. Condition. $\dfrac{2Er_1}{R+r_1+r_2}=E\Rightarrow R=r_1-r_2$.

Answer. Option A.

Point to note. $R=r_1-r_2$.

Q29JEE Main 2018 (8 Apr)Medium

Two batteries with e.m.f. $12$ V and $13$ V are connected in parallel across a load resistor of $10\ \Omega$. The internal resistances of the two batteries are $1\ \Omega$ and $2\ \Omega$ respectively. The voltage across the load lies between:

A$11.7$ V and $11.8$ V
B$11.6$ V and $11.7$ V
C$11.5$ V and $11.6$ V
D$11.4$ V and $11.5$ V

Answer and solution

Answer: Option C: $11.5$ V and $11.6$ V

Concept. Replace the two parallel cells by one equivalent cell, then use a divider.

Step 1. Equivalent cell. $E=\dfrac{12/1+13/2}{1+1/2}=12.33$ V, $r=\dfrac23\ \Omega$.

Step 2. Load voltage. $12.33\times\dfrac{10}{10.67}=11.56$ V.

Answer. Between $11.5$ and $11.6$ V (Option C).

Point to note. Equivalent cell first.

Q30JEE Main 2020 (4 Sep, Shift 1)Easy

A battery of $3.0\ \text{V}$ is connected to a resistor dissipating $0.5\ \text{W}$ of power. If the terminal voltage of the battery is $2.5\ \text{V}$, the power dissipated within the internal resistance is:

A$0.50\ \text{W}$
B$0.072\ \text{W}$
C$0.10\ \text{W}$
D$0.125\ \text{W}$

Answer and solution

Answer: Option C: $0.10\ \text{W}$

Concept. The lost voltage $E-V$ is across the internal resistance; the same current flows through it.

Step 1. Current. $\dfrac{0.5}{2.5}=0.2$ A.

Step 2. Internal power. $(3-2.5)\times0.2=0.1$ W.

Answer. Option C.

Point to note. Internal drop $\times I$.

Q31JEE Main 2019 (10 Apr, Shift 1)Medium

Two cells, each of emf $1.5\ \text{V}$ and internal resistance $r$, are connected in series with a $2\ \Omega$ resistor and a parallel combination of $15\ \Omega$ and $10\ \Omega$ resistors. An ideal voltmeter connected across the $10\ \Omega$ resistance reads $2\ \text{V}$. The internal resistance $r$ of each cell is:

A$0\ \Omega$
B$1.5\ \Omega$
C$0.5\ \Omega$
D$1\ \Omega$

Answer and solution

Answer: Option C: $0.5\ \Omega$

Concept. The voltmeter reading gives the current in the parallel pair; then $2E=I(R_{\text{total}}+2r)$.

Step 1. Parallel. $15\parallel10=6\ \Omega$; current $\dfrac26=\dfrac13$ A.

Step 2. Loop. $3=\dfrac13(2+6+2r)\Rightarrow r=0.5\ \Omega$.

Answer. Option C.

Point to note. Reading gives $I$.

Q32JEE Main 2024 (29 Jan, Shift 1)Hard

A $16\ \Omega$ wire is bent to form a square loop. A $9\ \text{V}$ battery with internal resistance $1\ \Omega$ is connected across one of its sides. If a $4\ \mu\text{F}$ capacitor is connected across one of its diagonals, the energy stored by the capacitor will be $\dfrac{x}{2}\ \mu\text{J}$, where $x =$ ______.

Numerical answer type

Answer and solution

Answer: 81

Concept. In steady state no current flows into the capacitor, so it just measures the voltage between two corners of the loop.

Step 1. Circuit. Sides $4\ \Omega$; the side across the battery is parallel with the other three ($12\ \Omega$): $3\ \Omega$. Current $\dfrac{9}{3+1}=2.25$ A; voltage $6.75$ V.

Step 2. Diagonal. Along the $12\ \Omega$ path the current is $0.5625$ A; two sides take $0.5625\times8=4.5$ V.

Step 3. Energy. $\tfrac12\times4\times10^{-6}\times4.5^2=40.5\ \mu$J $=\dfrac{81}{2}\ \mu$J.

Answer. $x=81$.

Point to note. Capacitor reads diagonal voltage.

Topic 5 · Q33–Q38

Wheatstone bridge and meter bridge

25 of the 256 JEE PYQs in the bank test this as their main idea

Look first

Null point: no current in G.

Quick notes

$$\text{balanced: }\frac PQ=\frac RS,\qquad \text{meter bridge: }\frac{R_{\text{left}}}{R_{\text{right}}}=\frac{l}{100-l}$$

At balance the galvanometer branch carries no current and can be removed. Swapping the cell and galvanometer keeps the balance.

Interchanging the two resistors moves the null point to $100-l$.
Shunting the right resistor (smaller) moves the null point to the right.
Balance does not depend on the wire's resistance per cm or thickness (if uniform).
Most sensitive when all four arms are of similar size and the null point is near the middle.
Cyclic network $P,Q,R,S$: balanced when $PR=QS$.
End corrections: measure with resistors swapped and average.

Trap. Measure $l$ from the end on the side of the resistor you call "left".

Practice: 6 questions

Q33JEE Main 2024 (1 Feb, Shift 2)Medium

In a metre bridge, when a resistance in the left gap is $2\ \Omega$ and an unknown resistance is in the right gap, the balance length is found to be $40\ \text{cm}$. On shunting the unknown resistance with $2\ \Omega$, the balance length changes by

A$22.5\ \text{cm}$
B$20\ \text{cm}$
C$62.5\ \text{cm}$
D$65\ \text{cm}$

Answer and solution

Answer: Option A: $22.5\ \text{cm}$

Concept. Meter bridge balance: $\dfrac{R_{\text{left}}}{R_{\text{right}}}=\dfrac{l}{100-l}$.

Step 1. Unknown. $\dfrac{2}{X}=\dfrac{40}{60}\Rightarrow X=3\ \Omega$.

Step 2. Shunted. $3\parallel2=1.2\ \Omega$; $\dfrac{2}{1.2}=\dfrac{l}{100-l}\Rightarrow l=62.5$ cm.

Step 3. Shift. $62.5-40=22.5$ cm.

Answer. Option A.

Point to note. $l/(100-l)$.

Q34JEE Main 2023 (29 Jan, Shift 1)Medium

In a metre bridge experiment the balance point is obtained if the gaps are closed by $2\ \Omega$ and $3\ \Omega$. A shunt of $X\ \Omega$ is added to the $3\ \Omega$ resistor to shift the balancing point by $22.5\ \text{cm}$. The value of $X$ is ______.

Numerical answer type

Answer and solution

Answer: 2

Concept. Same balance condition. Shunting the right resistor lowers it, so the null point moves right.

Step 1. First balance. $\dfrac23=\dfrac{l}{100-l}\Rightarrow l=40$ cm.

Step 2. New right resistance. At $62.5$ cm: $\dfrac{2}{R}=\dfrac{62.5}{37.5}\Rightarrow R=1.2\ \Omega$.

Step 3. Shunt. $\dfrac{3X}{3+X}=1.2\Rightarrow X=2\ \Omega$.

Answer. $2$.

Point to note. Shunt lowers right arm.

Q35JEE Main 2023 (29 Jan, Shift 2)Medium

When two resistances $R_1$ and $R_2$ connected in series and introduced into the left gap of a meter bridge and a resistance of $10\ \Omega$ is introduced into the right gap, a null point is found at $60\ \text{cm}$ from left side. When $R_1$ and $R_2$ are connected in parallel and introduced into the left gap, a resistance of $3\ \Omega$ is introduced into the right gap to get null point at $40\ \text{cm}$ from left end. The product of $R_1R_2$ is ______ $\Omega^2$.

Numerical answer type

Answer and solution

Answer: 30

Concept. Two balances give the series sum and the parallel value; their product gives $R_1R_2$.

Step 1. Series. $\dfrac{R_1+R_2}{10}=\dfrac{60}{40}\Rightarrow R_1+R_2=15$.

Step 2. Parallel. $\dfrac{R_p}{3}=\dfrac{40}{60}\Rightarrow R_p=2$.

Step 3. Product. $R_1R_2=R_p(R_1+R_2)=30\ \Omega^2$.

Answer. $30$.

Point to note. $R_1R_2=R_p(R_1+R_2)$.

Q36JEE Main 2020 (8 Jan, Shift 1)Medium

Four resistances of $15\ \Omega$, $12\ \Omega$, $4\ \Omega$ and $10\ \Omega$ respectively, in cyclic order, form a Wheatstone's network. The resistance that is to be connected in parallel with the resistance of $10\ \Omega$ to balance the network is ______ $\Omega$.

Numerical answer type

Answer and solution

Answer: 10

Concept. A Wheatstone network $P,Q,R,S$ in cyclic order balances when the products of opposite arms are equal: $PR=QS$.

Step 1. Condition. $15\times4=12\times S\Rightarrow S=5\ \Omega$.

Step 2. Shunt on $10\ \Omega$. $\dfrac{10x}{10+x}=5\Rightarrow x=10\ \Omega$.

Answer. $10$.

Point to note. $PR=QS$.

Q37JEE Main 2018 (8 Apr)Medium

On interchanging the resistances, the balance point of a meter bridge shifts to the left by $10$ cm. The resistance of their series combination is $1\ \text{k}\Omega$. How much was the resistance on the left slot before interchanging the resistances?

A$910\ \Omega$
B$990\ \Omega$
C$505\ \Omega$
D$550\ \Omega$

Answer and solution

Answer: Option D: $550\ \Omega$

Concept. Before: $\dfrac XY=\dfrac{l}{100-l}$. After swapping: the balance moves to the mirror position $100-l$. A shift of $10$ cm to the left means $100-l=l-10$.

Step 1. Balance length. $l=55$ cm.

Step 2. Ratio. $\dfrac XY=\dfrac{55}{45}$; with $X+Y=1000\ \Omega$: $X=550\ \Omega$.

Answer. Option D.

Point to note. Swap: $l\to100-l$.

Q38JEE Main 2024 (27 Jan, Shift 1)Medium

A wire of length $10\ \text{cm}$ and radius $\sqrt7\times10^{-4}\ \text{m}$ is connected across the right gap of a meter bridge. When a resistance of $4.5\ \Omega$ is connected on the left gap by using a resistance box, the balance length is found to be at $60\ \text{cm}$ from the left end. If the resistivity of the wire is $R\times10^{-7}\ \Omega\,\text{m}$, then value of $R$ is:

Answer and solution

Answer: Option C: 66

Concept. Find the wire's resistance from the balance, then $\rho=\dfrac{RA}{l}$.

Step 1. Resistance. $\dfrac{4.5}{X}=\dfrac{60}{40}\Rightarrow X=3\ \Omega$.

Step 2. Resistivity. $\rho=\dfrac{3\times\frac{22}{7}\times7\times10^{-8}}{0.1}=66\times10^{-7}\ \Omega$ m.

Answer. Option C.

Point to note. $\rho=RA/l$.

Topic 6 · Q39–Q43

Potentiometer

21 of the 256 JEE PYQs in the bank test this as their main idea

Look first

Balance compares voltages without drawing current.

Quick notes

$$E=kl,\qquad k=\frac{V_{\text{wire}}}{L}=\frac{I R_{\text{wire}}}{L},\qquad r=R\left(\frac{l_1}{l_2}-1\right)$$

At balance no current is drawn from the cell being tested, so the potentiometer measures true emf. $l_1$: cell open; $l_2$: cell shunted by $R$.

Comparing emfs: $\dfrac{E_1}{E_2}=\dfrac{l_1}{l_2}$.
Primary current $I=\dfrac{E_0}{R_{\text{wire}}+R_{\text{series}}+r_0}$; gradient $k=I\times$ resistance per length.
Shunted cell: balances $V=\dfrac{ER}{R+r}$, so $l\propto\dfrac{R}{R+r}$.
Sensitivity rises as $k$ falls: longer wire or smaller current.
No balance if the driver emf is less than the emf being measured.

Trap. Internal resistance of the tested cell does not affect the balance without a shunt.

Practice: 5 questions

Q39JEE Main 2026 (28 Jan, Shift 1)Medium

In the potentiometer, when the cell in the secondary circuit is shunted with $4\Omega$ resistance, the balance is obtained at the length 120 cm of wire. Now when the same cell is shunted with $12\Omega$ resistance, the balance is shifted to a length of 180 cm . The internal resistance of cell is ______ $\Omega$

Answer and solution

Answer: Option A: 4

Concept. A shunted cell supplies a current, so its terminal voltage $V=\dfrac{ER}{R+r}$ is what the potentiometer balances: $V\propto l$.

Step 1. Ratio. $\dfrac{120}{180}=\dfrac{4/(4+r)}{12/(12+r)}=\dfrac{4(12+r)}{12(4+r)}$.

Step 2. Solve. $2\cdot12(4+r)=3\cdot4(12+r)\Rightarrow96+24r=144+12r\Rightarrow r=4\ \Omega$.

Answer. Option A.

Point to note. $l\propto R/(R+r)$.

Q40JEE Main 2022 (27 Jun, Shift 1)Medium

A cell, shunted by a $8\ \Omega$ resistance, is balanced across a potentiometer wire of length $3$ m. The balancing length is $2$ m when the cell is shunted by $4\ \Omega$ resistance. The value of internal resistance of the cell will be ______ $\Omega$.

Numerical answer type

Answer and solution

Answer: 8

Concept. Same method as above.

Step 1. Ratio. $\dfrac32=\dfrac{8(4+r)}{4(8+r)}$.

Step 2. Solve. $12(8+r)=16(4+r)\Rightarrow r=8\ \Omega$.

Answer. $8$.

Point to note. $l\propto R/(R+r)$.

Q41JEE Main 2018 (8 Apr)Easy

In a potentiometer experiment, it is found that no current passes through the galvanometer when the terminals of the cell are connected across $52$ cm of the potentiometer wire. If the cell is shunted by a resistance of $5\ \Omega$, a balance is found when the cell is connected across $40$ cm of the wire. Find the internal resistance of the cell.

A$2.5\ \Omega$
B$1\ \Omega$
C$1.5\ \Omega$
D$2\ \Omega$

Answer and solution

Answer: Option C: $1.5\ \Omega$

Concept. Open cell balances at $l_1$ (its emf); shunted by $R$ it balances at $l_2$. Then $r=R\left(\dfrac{l_1}{l_2}-1\right)$.

Step 1. Apply. $5\left(\dfrac{52}{40}-1\right)=1.5\ \Omega$.

Answer. Option C.

Point to note. $r=R(l_1/l_2-1)$.

Q42JEE Main 2022 (26 Jul, Shift 2)Medium

A potentiometer wire of length $300\ \text{cm}$ is connected in series with a resistance $780\ \Omega$ and a standard cell of emf $4\ \text{V}$. A constant current flows through the potentiometer wire. The length of the null point for a cell of emf $20\ \text{mV}$ is found to be $60\ \text{cm}$. The resistance of the potentiometer wire is ______ $\Omega$.

Numerical answer type

Answer and solution

Answer: 20

Concept. Potential gradient $=\dfrac{V_{\text{wire}}}{L}$. Find the wire's voltage from the null point, then the current.

Step 1. Wire voltage. $20$ mV over $60$ cm, so $100$ mV over $300$ cm.

Step 2. Divider. $\dfrac{4R_w}{780+R_w}=0.1\Rightarrow3.9R_w=78$.

Answer. $R_w=20\ \Omega$.

Point to note. Wire voltage from null point.

Q43JEE Main 2020 (8 Jan, Shift 1)Medium

The length of a potentiometer wire is $1200$ cm and it carries a current of $60$ mA. For a cell of emf $5$ V and internal resistance of $20\ \Omega$, the null point on it is found to be at $1000$ cm. The resistance of the whole wire is:

A$80\ \Omega$
B$120\ \Omega$
C$60\ \Omega$
D$100\ \Omega$

Answer and solution

Answer: Option D: $100\ \Omega$

Concept. At balance no current flows in the cell, so its full emf $5$ V matches the wire voltage over $1000$ cm.

Step 1. Gradient. $\dfrac{5}{1000}$ V/cm, so the whole wire has $6$ V.

Step 2. Resistance. $\dfrac{6}{0.06}=100\ \Omega$.

Answer. Option D. (Internal resistance plays no part at balance.)

Point to note. No current at balance.

Topic 7 · Q44–Q50

Electrical power and heating

26 of the 256 JEE PYQs in the bank test this as their main idea

Look first

Choose I²R or V²/R by what is common.

Quick notes

$$P=VI=I^2R=\frac{V^2}{R},\qquad H=I^2Rt,\qquad R_{\text{bulb}}=\frac{V_{\text{rated}}^2}{P_{\text{rated}}}$$

Use $I^2R$ when the current is common (series) and $V^2/R$ when the voltage is common (parallel, same supply).

Bulb on a different supply: $P=P_{\text{rated}}\left(\dfrac{V}{V_{\text{rated}}}\right)^2$.
Bulbs in series: the one with the lower power rating (higher $R$) glows brighter; total $\dfrac{1}{P}=\sum\dfrac1{P_i}$.
Bulbs in parallel: powers add.
Kettle on the same supply: time $\propto R\propto$ element length.
Transmission efficiency: $\dfrac{P}{P+I^2R_{\text{line}}}$.
1 kWh $=3.6\times10^6$ J; 1 cal $=4.2$ J.

Trap. A 20% drop in current is a 36% drop in power.

Practice: 7 questions

Q44JEE Main 2019 (12 Jan, Shift 1)Easy

Two electric bulbs, rated at $(25\ \text{W}, 220\ \text{V})$ and $(100\ \text{W}, 220\ \text{V})$, are connected in series across a $220$ V voltage source. If the $25$ W and $100$ W bulbs draw powers $P_1$ and $P_2$ respectively, then:

A$P_1 = 4$ W, $P_2 = 16$ W
B$P_1 = 9$ W, $P_2 = 16$ W
C$P_1 = 16$ W, $P_2 = 9$ W
D$P_1 = 16$ W, $P_2 = 4$ W

Answer and solution

Answer: Option D: $P_1 = 16$ W, $P_2 = 4$ W

Concept. Ratings give resistances, $R=\dfrac{V^2}{P}$. In series the current is the same, so $P\propto R$.

Step 1. Resistances. $R_{25}=4R_{100}$.

Step 2. Current. $I=\dfrac{220}{5R_{100}}$.

Step 3. Powers. $P_{100}=I^2R_{100}=\dfrac{100}{25}=4$ W; $P_{25}=4\times4=16$ W.

Answer. Option D. (The lower-rated bulb glows brighter in series.)

Point to note. Series: $P\propto R$.

Q45JEE Main 2021 (26 Aug, Shift 2)Easy

An electric bulb of $500$ W at $100$ V is used in a circuit having a $200$ V supply. Calculate the resistance $R$ to be connected in series with the bulb so that the power delivered by the bulb is $500$ W.

A$30\ \Omega$
B$5\ \Omega$
C$20\ \Omega$
D$10\ \Omega$

Answer and solution

Answer: Option C: $20\ \Omega$

Concept. The bulb must still get its rated $100$ V and current; the series resistor drops the rest.

Step 1. Bulb. $I=\dfrac{500}{100}=5$ A.

Step 2. Resistor. $R=\dfrac{200-100}{5}=20\ \Omega$.

Answer. Option C.

Point to note. Drop the extra voltage.

Q46JEE Main 2020 (5 Sep, Shift 1)Medium

An electrical power line, having a total resistance of $2\ \Omega$, delivers $1\ \text{kW}$ at $220\ \text{V}$. The efficiency of the transmission line is approximately:

A$72\%$
B$91\%$
C$85\%$
D$96\%$

Answer and solution

Answer: Option D: $96\%$

Concept. Line loss $I^2R$; efficiency $=\dfrac{P_{\text{delivered}}}{P_{\text{delivered}}+P_{\text{loss}}}$.

Step 1. Current. $\dfrac{1000}{220}=4.55$ A.

Step 2. Loss. $4.55^2\times2=41.3$ W.

Step 3. Efficiency. $\dfrac{1000}{1041.3}=96\%$.

Answer. Option D.

Point to note. $P/(P+I^2R)$.

Q47JEE Main 2024 (8 Apr, Shift 2)Easy

Water boils in an electric kettle in 20 minutes after being switched on. Using the same main supply, the length of the heating element should be ______ to ______ times of its initial length if the water is to be boiled in 15 minutes.

Adecreased, $3/4$
Bincreased, $4/3$
Cdecreased, $4/3$
Dincreased, $3/4$

Answer and solution

Answer: Option A: decreased, $3/4$

Concept. Same supply: $P=\dfrac{V^2}{R}$, and the heat needed is fixed, so time $\propto R\propto$ length.

Step 1. Ratio. $\dfrac{15}{20}=\dfrac34$.

Answer. Decreased to $\dfrac34$ (Option A).

Point to note. Time $\propto R\propto l$.

Q48JEE Main 2023 (13 Apr, Shift 1)Easy

A potential $V_0$ is applied across a uniform wire of resistance $R$. The power dissipation is $P_1$. The wire is then cut into two equal halves and a potential of $V_0$ is applied across the length of each half. The total power dissipation across two wires is $P_2$. The ratio of $P_2:P_1$ is $\sqrt x:1$. The value of $x$ is ______.

Numerical answer type

Answer and solution

Answer: 16

Concept. At fixed voltage $P=\dfrac{V^2}{R}$. Each half has $\dfrac R2$ and gets the full $V_0$.

Step 1. Total. $2\times\dfrac{V_0^2}{R/2}=4P_1$.

Step 2. Form. $4=\sqrt{16}$.

Answer. $x=16$.

Point to note. $V^2/(R/2)$ twice.

Q49JEE Main 2026 (2 Apr, Shift 2)Easy

Two resistors of $200\ \Omega$ and $400\ \Omega$ are connected in series with a battery of $100$ V. A bulb rated at $200$ V, $100$ W is connected across the $400\ \Omega$ resistance. The potential drop across the bulb is ______ V.

A$25$
B$50$
C$66.6$
D$100$

Answer and solution

Answer: Option B: $50$

Concept. The bulb is a resistor, $R=\dfrac{V^2}{P}=400\ \Omega$, in parallel with the $400\ \Omega$ resistor.

Step 1. Parallel. $200\ \Omega$, in series with $200\ \Omega$.

Step 2. Voltage. Half of $100$ V $=50$ V.

Answer. Option B.

Point to note. Bulb as $400\ \Omega$.

Q50JEE Main 2019 (11 Jan, Shift 1)Easy

Two equal resistances when connected in series to a battery, consume electric power of $60$ W. If these resistances are now connected in parallel combination to the same battery, the electric power consumed will be:

A$60$ W
B$240$ W
C$120$ W
D$30$ W

Answer and solution

Answer: Option B: $240$ W

Concept. Same battery: $P=\dfrac{V^2}{R_{\text{eq}}}$. Series $2R$, parallel $\dfrac R2$: four times smaller resistance.

Step 1. Power. $4\times60=240$ W.

Answer. Option B.

Point to note. Series to parallel: $\times4$.

Before the exam

Last-day checklist

$I=neAv_d$; $J=\sigma E$; $\rho=m/(ne^2\tau)$.
Stretch: $l\times n$ gives $R\times n^2$; $r/n$ gives $R\times n^4$.
$R_t=R_0(1+\alpha t)$; colour code BBROYGBVGW.
$n$-gon edge $(n-1)R/n^2$; circle arc $f(1-f)R$.
$V=E\mp Ir$; max power at $R=r$; parallel cells weighted by $1/r$.
Bridge $P/Q=R/S$; meter bridge $l/(100-l)$.
Potentiometer $E=kl$; $r=R(l_1/l_2-1)$.
$P=V^2/R$ parallel, $I^2R$ series; bulb $R=V^2/P$.

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Current Electricity

How to study this chapter

Where the marks come from

Contents

Electric current, drift speed and current density

Look first

Quick notes

Practice: 6 questions

Resistance, resistivity and temperature

Quick notes

Practice: 9 questions

Resistor combinations and meters

Quick notes

Practice: 8 questions

Cells, internal resistance and Kirchhoff's laws

Look first

Quick notes

Practice: 9 questions

Wheatstone bridge and meter bridge

Look first

Quick notes

Practice: 6 questions

Potentiometer

Look first

Quick notes

Practice: 5 questions

Electrical power and heating

Look first

Quick notes

Practice: 7 questions

Last-day checklist

Every chapter, in this same format

Complete Pack

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