Q159 / 170 · Manipal (MET) 2019 · Physics
Consider four circuits shown in the figure below. In which circuit power dissipated is greatèst?
(Neglect the internal resistance of the power supply).
- Correct answerA
- B
- C
- D
Solution
Let each resistance is $R$
In choice (a) $P=\frac{E^2}{R_{\mathrm{eq}}}$ $\Rightarrow \quad P_1=\frac{E^2}{\frac{R}{2}}=\frac{2 E^2}{R}$
In choice (b) $\begin{aligned} R_{\mathrm{eq}} & =R+R=2 R \\ P_2 & =\frac{E^2}{2 R}=0.5 \frac{E^2}{R} \end{aligned}$
In choice (c) $R_{\mathrm{eq}}=\frac{R}{2}+R=\frac{3 R}{2}$ $P_3=\frac{2 E^2}{3 R}=0.6 \frac{E^2}{R}$
In choice (d) $\begin{aligned} R & =\frac{2 R \times R}{2 R+R}=\frac{2 R}{3} \\ P_4 & =\frac{3 E^2}{2 R}=1.5 \frac{E^2}{R} \end{aligned}$
So, $P_1$ is maximum.
In choice (a) $P=\frac{E^2}{R_{\mathrm{eq}}}$ $\Rightarrow \quad P_1=\frac{E^2}{\frac{R}{2}}=\frac{2 E^2}{R}$
In choice (b) $\begin{aligned} R_{\mathrm{eq}} & =R+R=2 R \\ P_2 & =\frac{E^2}{2 R}=0.5 \frac{E^2}{R} \end{aligned}$
In choice (c) $R_{\mathrm{eq}}=\frac{R}{2}+R=\frac{3 R}{2}$ $P_3=\frac{2 E^2}{3 R}=0.6 \frac{E^2}{R}$
In choice (d) $\begin{aligned} R & =\frac{2 R \times R}{2 R+R}=\frac{2 R}{3} \\ P_4 & =\frac{3 E^2}{2 R}=1.5 \frac{E^2}{R} \end{aligned}$
So, $P_1$ is maximum.
Hence, the correct answer is option: 'A'.