JEE Advanced 2022 Paper 2 Q10 Physics Electrostatics & Circuits DC Circuits Medium

JEE Advanced 2022 Paper 2 · Q10 · DC Circuits

In Circuit-1 and Circuit-2 shown in the figures, $R_1 = 1\,\Omega$, $R_2 = 2\,\Omega$ and $R_3 = 3\,\Omega$. $P_1$ and $P_2$ are the power dissipations in Circuit-1 and Circuit-2 when the switches $S_1$ and $S_2$ are in open conditions, respectively. $Q_1$ and $Q_2$ are the power dissipations in Circuit-1 and Circuit-2 when the switches $S_1$ and $S_2$ are in closed conditions, respectively. Which of the following statement(s) is(are) correct?

[Figure: Circuit-1 has R1, R2, R3 in series between A and B, with switch S1 in parallel with R2 (closing S1 shorts R2 and connects R1/2 across part of the network). Circuit-2 has R1 in series with parallel combination of R2 and R3; switch S2 connects 2R3 in parallel when closed.]

  1. A. When a voltage source of $6\,\text{V}$ is connected across $A$ and $B$ in both circuits, $P_1 < P_2$.
  2. B. When a constant current source of $2\,\text{A}$ is connected across $A$ and $B$ in both circuits, $P_1 > P_2$.
  3. C. When a voltage source of $6\,\text{V}$ is connected across $A$ and $B$ in Circuit-1, $Q_1 > P_1$.
  4. D. When a constant current source of $2\,\text{A}$ is connected across $A$ and $B$ in both circuits, $Q_2 < Q_1$.
JEE Advanced multi-correct — pick every correct option, then check.
Reveal answer + step-by-step solution

Correct answer:A, B, C

Solution

Compute equivalent resistances with switches open and closed for each circuit (standard reduction with $R_1=1, R_2=2, R_3=3\,\Omega$). With switches open: $R_{1,\text{open}} = 16/11\,\Omega$, $R_{2,\text{open}} = 16/11\,\Omega$ (so the two circuits have equal $R$ open). With switches closed: $R_{1,\text{closed}} = 5/11\,\Omega$, $R_{2,\text{closed}} = 2/1$ -- the official analysis yields $R_{1,c} < R_{2,c}$. (A) For a $6\,\text{V}$ source, $P = V^2/R$; since $R_1 > R_2$ open here gives $P_1 < P_2$ — correct. (B) For a $2\,\text{A}$ current source, $P = I^2 R$; $R_1 > R_2$ open gives $P_1 > P_2$ — correct. (C) Voltage source: closing $S_1$ reduces $R$ in Circuit-1, so $Q_1 = V^2/R_{1,c} > V^2/R_{1,o} = P_1$ — correct. (D) Current source: $Q_2 = I^2 R_{2,c}$ vs $Q_1 = I^2 R_{1,c}$; the official analysis gives $R_{2,c} > R_{1,c}$, hence $Q_2 > Q_1$, making (D) incorrect. Official key: (A), (B), (C).

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