Q 12-03-098JEE MainJEE Main 2024 (9 Apr, Shift 1)Hard
The current flowing through the $1\ \Omega$ resistor is $\dfrac{n}{10}\ \text{A}$. The value of $n$ is ______.
Numerical value type. Enter your answer.
Answer: 25
Take $V_D = 0$; the 5 V cell gives $V_B = 5\ \text{V}$. Let the potentials of A and C be $V_A$ and $V_C$. The 10 V cell raises the potential towards A, so the current in the $1\ \Omega$ branch (C to A) is $I_1 = V_C + 10 - V_A$.
KCL at A: $I_1 = \dfrac{V_A - 5}{4} + \dfrac{V_A}{4}$
KCL at C: $I_1 + \dfrac{V_C - 5}{2} + \dfrac{V_C}{2} = 0$
Solving: $V_C = 0$, $V_A = 7.5\ \text{V}$, so
$$I_1 = 0 + 10 - 7.5 = 2.5\ \text{A} = \frac{25}{10}\ \text{A} \;\Rightarrow\; n = 25$$
(Check at A: $\dfrac{2.5}{4} + \dfrac{7.5}{4} = 2.5\ \text{A}$.)
Solution by Sreeraj P, M.Sc Physics