Question

Two resistances 30 $\Omega$ and 40 $\Omega$ are connected in left and right gaps of the meterbridge respectively. The resistance of bridge wire is 7 $\Omega$. The battery of e.m.f. 7 volt is connected to two ends of the wire. Then the current through cell will be (Internal resistance of battery is neglected)

• A. 0.4 A
• B. 0.5 A
• C. 1.1 A
• D. 0.9 A

Answer 1

Answer: (C)

1.1 A

Answer 2

We can model the circuit as two parallel branches between points A and B:

1. Branch 1: The meterbridge wire with resistance $R_1 = 7\,\Omega$.

2. Branch 2: The series combination of the two resistors, giving

   $$R_2 = 30\,\Omega + 40\,\Omega = 70\,\Omega.$$

The equivalent resistance for two resistors in parallel is:

$$R_{\text{eq}} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{7 \times 70}{7 + 70} = \frac{490}{77} \approx 6.3636\,\Omega.$$

Now using Ohm’s law with emf $V = 7\,\text{V}$:

$$I = \frac{V}{R_{\text{eq}}} \approx \frac{7}{6.3636} \approx 1.1\,\text{A}.$$

Thus, the current through the cell is approximately 1.1 A.