Thank you for reporting, we will resolve it shortly
Q.
In the given circuit, $R_1 = 10 \Omega, R_2 = 6\Omega$ and $E = 10\, V$. Then reading of $A_2$ is:
Current Electricity
Solution:
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.
