Question

If $R_1 = R_2 = R$ and $R_3 = 1.1R_4$ in the bridge circuit shown in figure, then the reading in the ideal voltmeter connected between $a$ and $b$ is:

• A. 0.238 V
• B. 0.138 V
• C. -0.238 V
• D. 1 V

Hint:

In bridge circuits, always pay attention to polarity markings. Use voltage division to find $V_a$ and $V_b$, and subtract accordingly.

Answer 1

The Correct Option is C

This is a classic Wheatstone bridge problem. We are to find the voltage difference $V_{ab} = V_a - V_b$.
Given:
- $R_1 = R_2 = R$
- $R_3 = 1.1R$
- $R_4 = R$
- Supply voltage = 10V across the vertical leg
Let’s calculate the potentials at points $a$ and $b$ using voltage division.
Left branch (with $R_1$ and $R_2$):
\[ V_a = \frac{R_2}{R_1 + R_2} \cdot 10V = \frac{R}{R + R} \cdot 10 = \frac{1}{2} \cdot 10 = 5V \]
Right branch (with $R_3 = 1.1R$ and $R_4 = R$):
\[ V_b = \frac{R_4}{R_3 + R_4} \cdot 10V = \frac{R}{1.1R + R} \cdot 10 = \frac{1}{2.1} \cdot 10 \approx 4.762V \]
Now calculate the voltmeter reading: \[ V_{ab} = V_a - V_b = 5V - 4.762V = 0.238V \]
However, the question asks for voltmeter connected from $a$ to $b$, and the polarity marked in the figure shows that $a$ is at the positive terminal of the voltmeter.
Since $V_a>V_b$, the actual reading would be positive, but the meter shows $+$ at $b$ and $-$ at $a$, indicating reverse polarity.
Hence, the voltmeter would read: \[ \boxed{-0.238\ \text{V}} \]