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QJEEAdv22Ph31da: JEE Advanced 2022 Physics - Class 12 JEEAdv22Ph31da Solution: In Circuit-1 and Circuit- 2 shown in the figures, and . and are …

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

\mathrm{S}_{1}

and

\mathrm{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

\mathrm{S}_{1}

and

\mathrm{S}_{2}

are in closed conditions, respectively.

JEE Advanced 2022 Paper 2 Online Physics - Current Electricity Question 4 English

JEE Advanced 2022 Paper 2 Online Physics - Current Electricity Question 4 English

Which of the following statement(s) is(are) correct?

Multiple Choice Options

A When a voltage source of $6 V$ is connected across $\mathrm{A}$ and $\mathrm{B}$ in both circuits, $P_{1} < P_{2}$ .
B When a constant current source of $2 \mathrm{Amp}$ is connected across A and B in both circuits, $P_{1}>P_{2}$ .
C When a voltage source of $6 V$ is connected across $\mathrm{A}$ and $\mathrm{B}$ in Circuit-1, $Q_{1}>P_{1}$ .
D When a constant current source of $2 \mathrm{Amp}$ is connected across A and $\mathrm{B}$ in both circuits, $Q_{2} < Q_{1}$ .

Step-by-Step Verified Solution

JEE Advanced 2022 Paper 2 Online Physics - Current Electricity Question 4 English Explanation 1

JEE Advanced 2022 Paper 2 Online Physics - Current Electricity Question 4 English Explanation 2

Case 1 : When

\mathrm{S}_1

and

\mathrm{S}_2

are open

\left(R_{e q}\right)_1=1+\frac{5 \times \frac{1}{2}}{5+\frac{1}{2}}=1+\frac{5}{11}=\frac{16}{11}

P_1=\frac{V^2}{(R_{e q})_1}=\frac{(6)^2}{16} \times 11=\frac{36 \times 11}{16}=24.75 \mathrm{~W}

\left(R_{e q}\right)_2=\frac{6}{11} \Omega

P_2=\frac{V^2}{(R_{e q})_2}=\frac{(6)^2}{6} \times 11=\frac{36 \times 11}{6}=66 \mathrm{~W}

P_2&gt;P_1

Option (A) is correct.

Case 2 : If I =

2 \mathrm{~A}

source is used in both circuits, then

P_1=i^2\left(R_{e q}\right)_1=(2)^2 \times \frac{16}{11}=\frac{64}{11}=5.818 \mathrm{~W}

P_2=i^2\left(R_{e q}\right)_2=(2)^2 \times \frac{6}{11}=\frac{24}{11}=2.1818 \mathrm{~W}

P_1&gt;P_2

Option (B) is correct.

Case 3 : For

Q_1

\begin{aligned}&amp; R_{e q}=\frac{5}{11} \Omega \\\\&amp; Q_1=\frac{V^2}{R_{e q}}=\frac{(6)^2}{\frac{5}{11}}=\frac{36 \times 11}{5}=79.2 \mathrm{~W} \\\\&amp; P_1=24.75 \mathrm{~W} \\\\&amp; Q_1&gt;P_1\end{aligned}

Option (C) is correct.

Case 4 : For option (D)

\begin{aligned}&amp; Q_1=i^2 R_{e q}=(2)^2 \times \frac{5}{11}=\frac{20}{11}=1.81 \mathrm{~W} \\\\&amp; Q_2=i^2 R_{e q}=(2)^2 \times \frac{1}{2}=\frac{4}{2}=2 \mathrm{~W} \\\\&amp; Q_2&gt;Q_1\end{aligned}

Option (D) is incorrect.

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