Question

A galvanometer has a coil of resistance \(100\,\Omega\) showing a full scale deflection at \(50\,\mu\text{A}\). The resistance that should be added to use it as an ammeter of range \(10\,\text{mA}\) is

• A. \(5\,\Omega\)
• B. \(5\times 10^{-2}\,\Omega\)
• C. \(0.5\,\Omega\)
• D. \(1\,\Omega\)

Hint:

To convert a galvanometer into an ammeter, connect a low resistance shunt in parallel: \[ S=\frac{I_gG}{I-I_g} \] where \(G\) is galvanometer resistance and \(I_g\) is full scale current.

Answer 1

The Correct Option is C

Step 1: Understand the conversion of a galvanometer to an ammeter.
A galvanometer measures small currents. To convert it to an ammeter with a larger range, we connect a low resistance (called a shunt $S$) in parallel with the galvanometer. The shunt bypasses most of the current so that the galvanometer only carries its safe maximum current $I_g$.
Step 2: Write down the given data.
Galvanometer coil resistance: $G = 100\,\Omega$. Full scale deflection current: $I_g = 50\,\mu\text{A} = 50 \times 10^{-6}\,\text{A}$. Desired ammeter range: $I = 10\,\text{mA} = 10 \times 10^{-3}\,\text{A}$.
Step 3: Apply the shunt formula.
Since the galvanometer and shunt are in parallel, the voltage across both is equal: \[ I_g G = (I - I_g) S \] Rearranging for $S$: \[ S = \frac{I_g G}{I - I_g} \]
Step 4: Substitute the values.
\[ S = \frac{(50 \times 10^{-6})(100)}{10 \times 10^{-3} - 50 \times 10^{-6}} = \frac{5 \times 10^{-3}}{9.95 \times 10^{-3}} \]
Step 5: Compute the shunt resistance.
\[ S = \frac{5}{9.95} \approx 0.503\,\Omega \approx 0.5\,\Omega \] The shunt is much smaller than $G = 100\,\Omega$, which makes sense because it must carry most of the current.
Step 6: State the final answer.
The resistance that should be added (as a shunt) is: \[ \boxed{0.5\,\Omega} \]