Q 12-03-125JEE MainJEE Main 2023 (24 Jan, Shift 1)Medium
As shown in the figure, a network of resistors is connected to a battery of $24\ \text{V}$ with an internal resistance of $3\ \Omega$. The currents through the resistors $R_4$ and $R_5$ are $I_4$ and $I_5$ respectively. The values of $I_4$ and $I_5$ are
Answer: (D) $I_4=\dfrac25\ \text{A}$ and $I_5=\dfrac85\ \text{A}$
$R_1\parallel R_2=1\ \Omega$, $R_4\parallel R_5=\dfrac{20\times5}{25}=4\ \Omega$.
Total resistance $=1+2+4+2+3=12\ \Omega$, so $I=\dfrac{24}{12}=2\ \text{A}$.
Voltage across $R_4\parallel R_5$: $2\times4=8\ \text{V}$. Hence $I_4=\dfrac{8}{20}=\dfrac25\ \text{A}$ and $I_5=\dfrac85\ \text{A}$.
Solution by Sreeraj P, M.Sc Physics