Question

In the given circuit, $R_1=10\Omega ,R_2=6\Omega$ and $E=10V$. Then reading of $A_2$ is:

• A. 1 amp
• B. 2 amp
• C. 1/4 amp
• D. 3 amp

Answer 1

Answer: (C)

Potential difference across $R_2$ resistances is zero, therefore current in three branches is zero, therefore current in two branch containing $R_1$ will be same, simplified circuit will be

Effective resistance of the circuit
$R_{\text{eff }}=\frac{40\times 40}{40+40}$
$=20\Omega$
Current through the circuit
$I=\frac{10}{20}=(1/2)$ amp. Hence
reading of $A_1=1/2$ amp.
Hence reading of $A_2=1/4$ amp.