Question

A galvanometer of resistance \(50\Omega\) is connected to a battery of \(4\,V\) along with a resistance of \(3950\Omega\) in series. A full-scale deflection of 30 divisions is obtained in the galvanometer. In order to reduce this deflection to 10 divisions, the resistance in series should be equal to:

• A. \(8950\Omega\)
• B. \(11950\Omega\)
• C. \(7000\Omega\)
• D. \(6000\Omega\)

Hint:

Galvanometer deflection \( \alpha\) current, so use ratio method instead of solving separately.

Answer 1

The Correct Option is B

Step 1: Relation between deflection and current.
Deflection in galvanometer is directly proportional to current.
\[ \theta \propto I \]

Step 2: Ratio of currents.
\[ \frac{I_1}{I_2} = \frac{30}{10} = 3 \]

Step 3: Initial current.
\[ I_1 = \frac{V}{R + R_g} = \frac{4}{3950 + 50} = \frac{4}{4000} \]

Step 4: New current expression.
Let new resistance be \(R'\).
\[ I_2 = \frac{4}{R' + 50} \]

Step 5: Use current ratio.
\[ \frac{I_1}{I_2} = \frac{R' + 50}{4000} = 3 \]

Step 6: Solve for \(R'\).
\[ R' + 50 = 12000 \] \[ R' = 11950 \Omega \]

Step 7: Final conclusion.
\[ \boxed{11950\Omega} \] Hence, correct answer is option (B).