Question

In Circuit-1 and Circuit-2 shown in the figures, $R_1=1\Omega ,R_2=2\Omega \text{and}{R}_3=3\Omega$. $P_1\text{and}{P}_2$ are the power dissipations in Circuit-1 and Circuit-2 when the switches $S_1$ and $S_2$ are in open conditions, respectively.
$Q_1\text{and}{Q}_2$ are the power dissipations in Circuit-1 and Circuit-2 when the switches $S_1\text{and}{S}_2$ are in closed conditions, respectively.




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

• A. When a constant current source of $2\text{Amp}$ is connected across $A$ and $B$ in both circuits, $Q_2<Q_1$.
• B. When a constant current source of $2\text{Amp}$ is connected across $A$ and $B$ in both circuits. $P_1>P_2$.
• C. When a voltage source of $6\text{V}$ is connected across $A$ and $B$ in Circuit-1, $Q_1>P_1$.
• D. When a voltage source of $6\text{V}$ is connected across $A$ and $B$ in both circuits, $P_1<P_2$.

Answer 1

Answer: (B), (C), (D)

When a voltage source of $6\text{V}$ is connected across $A$ and $B$ in both circuits, $P_1<P_2$.

$R_{eq}=\frac{5\times \frac{1}{2}}{5+\frac{1}{2}}+1=\frac{16}{11}\Omega$

$R_{eq}=\frac{5\times \frac{1}{2}}{5+\frac{1}{2}}=\frac{5}{11}\Omega$

$\frac{1}{R_{eq}}=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}\Rightarrow R_{eq}=\frac{6}{11}\Omega$

$\frac{1}{R_{eq}}=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=\frac{1}{2}\Omega$

For constant $V$, $P\propto \frac{1}{R}\Rightarrow P_1<P_2$
$\therefore$Options $A$ and $C$ are right.
For constant $I$, $P\propto R\Rightarrow P_2<P_1$
$\therefore$Option $B$ is right.