Q174 / 180 · COMEDK 2024 (Memory Based) · Physics
The current drawn from the battery by the network of resistors as shown in the figure is
- A1 A
- Correct answerB2 A
- C0.5 A
- D0 A
Solution
The given circuit is equivalent to given below,
Now, $\frac{1 \Omega}{2 \Omega}=\frac{2 \Omega}{4 \Omega}$
i.e.,
$\frac{P}{Q}=\frac{R}{S}$
The given circuit is a balanced Wheatstone bridge. The resistance of $5 \Omega$ in arm $B D$ is ineffective.
$\therefore$ Now circuit can be drawn as
$\therefore$ Equivalent resistance,
$\begin{aligned}<br/>R^{\prime} & =(1+2) \|(2+4) \\<br/>& =3 \| 6=\frac{3 \times 6}{3+6}=2 \Omega<br/>\end{aligned}$
$\therefore$ Current drawn from the battery,
$I=\frac{V}{R^{\prime}}=\frac{4}{2}=2 \mathrm{~A}$
Now, $\frac{1 \Omega}{2 \Omega}=\frac{2 \Omega}{4 \Omega}$
i.e.,
$\frac{P}{Q}=\frac{R}{S}$
The given circuit is a balanced Wheatstone bridge. The resistance of $5 \Omega$ in arm $B D$ is ineffective.
$\therefore$ Now circuit can be drawn as
$\therefore$ Equivalent resistance,
$\begin{aligned}<br/>R^{\prime} & =(1+2) \|(2+4) \\<br/>& =3 \| 6=\frac{3 \times 6}{3+6}=2 \Omega<br/>\end{aligned}$
$\therefore$ Current drawn from the battery,
$I=\frac{V}{R^{\prime}}=\frac{4}{2}=2 \mathrm{~A}$
Hence, the correct answer is option: 'B'.