Question

In the circuit shown, the galvanometer (G) has an internal resistance of $100 \Omega$. The galvanometer current $I_G$ is ________ $\mu A$ (rounded off to the nearest integer).

 

Hint:

Use Thevenin's theorem to simplify the bridge circuit seen by the galvanometer. Calculate the open-circuit voltage across the galvanometer terminals and the equivalent resistance looking into those terminals.

Answer 1

Step 1: Calculate the voltage at nodes B and D.
$V_B = 4 \times \frac{1010}{1000 + 1010} = 2.0199 V$
$V_D = 4 \times \frac{1000}{1000 + 1000} = 2 V$
Step 2: Calculate the Thevenin voltage $V_{TH}$.
$V_{TH} = V_B - V_D = 0.0199 V$
Step 3: Calculate the Thevenin resistance $R_{TH}$.
$R_{TH} = (1000 || 1010) + (1000 || 1000) = 502.4876 + 500 = 1002.4876 \Omega$
Step 4: Calculate the galvanometer current $I_G$.
$I_G = \frac{V_{TH}}{R_{TH} + R_G} = \frac{0.0199}{1002.4876 + 100} = \frac{0.0199}{1102.4876} = 1.805 \times 10^{-5} A$
Step 5: Convert to $\mu A$ and round off.
$I_G = 18.05 \mu A \approx 18 \mu A$