Q154 / 180 · COMEDK 2017 · Physics
In the circuit given $E=6.0 \mathrm{~V}, R_{1}=100 \Omega$ $R_{2}=R_{3}=50 \Omega$ and $R_{4}=75 \Omega$.
The equivalent resistance of the circuit (in $\Omega$ ) is
- A$11.875$
- B$26.31$
- Correct answerC$118.75$
- DNone of these
Solution
Given, $R_{1}=100 \Omega, R_{2}=R_{3}=50 \Omega, E=6 \mathrm{~V}$,
$R_{4}=75 \Omega$
As $R_{2}, R_{3}$ and $R_{4}$ are in parallel combination, so their equivalent resistance be
$\begin{aligned}
\frac{1}{R^{\prime}} &=\frac{1}{R_{2}}+\frac{1}{R_{3}}+\frac{1}{R_{4}} \\
&=\frac{1}{50}+\frac{1}{50}+\frac{1}{75}=\frac{3+3+2}{150}=\frac{8}{150}
\end{aligned}$
or $R^{\prime}=\frac{150}{8} \Omega$
Now, $R^{\prime}$ is in series with $R_{1}$, so
$R_{\text {net }}=R_{1}+R^{\prime}=100+\frac{150}{8}=\frac{950}{8}=118.75 \Omega$
Hence, the correct answer is option: 'C'.