Animated Solution for Physics - Current Electricity: 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?

Choosing the Right Power Formula

When dealing with power dissipation in electrical circuits, the most critical decision is selecting the appropriate formula based on the constraints of the problem. We have two primary formulas for power $P$ dissipated by a resistance $R$:

1. Voltage Source: When a constant voltage $V$ is applied across the terminals, we use $P=\frac{V^2}{R}$. In this scenario, power is inversely proportional to the equivalent resistance ($P\propto \frac{1}{R}$). A lower resistance draws more current, leading to higher power dissipation. 2. Current Source: When a constant current $I$ is forced through the circuit, we use $P=I^{2}R$. Here, power is directly proportional to the equivalent resistance ($P\propto R$). A higher resistance requires a larger voltage drop to maintain the same current, resulting in higher power dissipation.

Keeping these relationships in mind is the key to unlocking this problem.

Let's first examine the circuits when both switches, $S_1$ and $S_2$, are open.

For Circuit-1, the equivalent resistance is given as:

For Circuit-2, observing the diagram reveals that resistors $R_1$, $R_2$, and $R_3$ are connected purely in parallel across the terminals A and B. We can calculate its equivalent resistance $R_{C_2}$:

Now, let's evaluate the first two options:

Option (A): A constant voltage source is connected. We compare $P_1$ and $P_2$. Since $P\propto \frac{1}{R}$ and $R_{C_1}\left(\frac{16}{11}\right)>R_{C_2}\left(\frac{6}{11}\right)$, it strictly follows that $P_1<P_2$. Option (A) is correct.

Option (B): A constant current source is connected. We compare $P_1$ and $P_2$. Since $P\propto R$ and $R_{C_1}>R_{C_2}$, it follows that $P_1>P_2$. Option (B) is also correct.

Next, we analyze the circuits when the switches are closed.

For Circuit-1, closing $S_1$ alters the network topology, and the new equivalent resistance is given as:

For Circuit-2, closing $S_2$ introduces a new branch with resistance $2R_3=2(3)=6\Omega$ in parallel with the existing network. The new equivalent resistance $R_{C_2}^{\prime }$ is:

Now, let's evaluate the remaining options:

Option (C): We compare Circuit-1 in its closed state ($Q_1$) and open state ($P_1$) with a constant voltage source. Since $P\propto \frac{1}{R}$ and the resistance decreased upon closing the switch ($R_{C_1}^{\prime }<R_{C_1}$), the power dissipation must increase. Therefore, $Q_1>P_1$. Option (C) is correct.

Option (D): We compare both circuits in their closed states ($Q_1$ and $Q_2$) with a constant current source. Since $P\propto R$, we must compare $R_{C_1}^{\prime }$ and $R_{C_2}^{\prime }$. We have $R_{C_2}^{\prime }=0.5\Omega$ and $R_{C_1}^{\prime }=\frac{5}{11}\approx 0.45\Omega$. Because $R_{C_2}^{\prime }>R_{C_1}^{\prime }$, it must be that $Q_2>Q_1$. The statement claims $Q_2<Q_1$, which is false. Option (D) is incorrect.

By systematically applying the correct power formulas based on the source type and carefully tracking the equivalent resistances, we have determined that statements (A), (B), and (C) are correct.