Question

A galvanometer having a resistance of \(50\Omega\) is shunted by a wire of resistance \(10\Omega\). If the total current is \(2\,\text{A}\), the current passing through the shunt is:

• A. \(\frac{3}{5}\,\text{A}\)
• B. \(\frac{2}{5}\,\text{A}\)
• C. \(\frac{5}{6}\,\text{A}\)
• D. \(\frac{5}{3}\,\text{A}\)

Hint:

In a parallel combination, current divides inversely proportional to resistance. The smaller resistance branch carries greater current.

Answer 1

The Correct Option is D

Step 1: Identify the parallel combination.
The galvanometer resistance and shunt resistance are connected in parallel. Therefore, the potential difference across both is the same.

Step 2: Write given values.
\[ R_g = 50\Omega \]
\[ R_s = 10\Omega \]
\[ I = 2\,\text{A} \]

Step 3: Use current division rule.
For two resistances in parallel, current through shunt is:
\[ I_s = I \times \frac{R_g}{R_g + R_s} \]

Step 4: Substitute values.
\[ I_s = 2 \times \frac{50}{50 + 10} \]
\[ I_s = 2 \times \frac{50}{60} \]

Step 5: Simplify the fraction.
\[ I_s = 2 \times \frac{5}{6} \]
\[ I_s = \frac{10}{6} \]
\[ I_s = \frac{5}{3}\,\text{A} \]

Step 6: Check using current distribution logic.
Since the shunt resistance \(10\Omega\) is smaller than the galvanometer resistance \(50\Omega\), a larger current flows through the shunt. Therefore, \(\frac{5}{3}\,\text{A}\) is reasonable.

Step 7: Final answer.
\[ \boxed{\frac{5}{3}\,\text{A}} \]