Question

A galvanometer of resistance $27 \Omega$ is converted into an ammeter of range (0 – 10 mA) using a resistance of $3 \Omega$. The galvanometer will show full scale deflection for a current of about –

• A. 10 mA
• B. 100 mA
• C. 1 mA
• D. 3 mA

Hint:

Notice that the shunt resistance ($3 \Omega$) is exactly $1/9$th of the galvanometer resistance ($27 \Omega$). This instantly means the shunt will greedily take $9$ times more current than the galvanometer. If total current is $10$ parts, galvanometer takes exactly $1$ part.

Answer 1

The Correct Option is C

Concept:
• A sensitive galvanometer is effectively and safely converted into a much higher-range ammeter by connecting a very low resistance resistor, called a shunt, strictly in parallel with it.

• Because the galvanometer coil and the shunt resistor are firmly connected in parallel, the electrical potential difference (voltage drop) across both distinct pathways must be absolutely identical.

• This creates a neat division of the total incoming current $I$: a very small, safe fraction $I_g$ flows through the delicate galvanometer, while the massive remainder $(I - I_g)$ bypasses it securely through the rugged shunt.

Step 1: Extract the known variables from the problem
The internal electrical resistance of the bare galvanometer coil is strictly $R_g = 27 \Omega$.
The tiny parallel shunt resistance creatively used for the conversion is firmly $S = 3 \Omega$.
The maximum target measurable range of the newly constructed ammeter is the total current $I = 10 \text{ mA}$.
The objective is to accurately find the specific current $I_g$ that physically forces the galvanometer to show its maximum full-scale deflection.

Step 2: Establish the foundational parallel voltage equation
Since the two resistors are strictly parallel, we systematically equate their voltage drops using Ohm's Law ($V = IR$):
\[ V_{\text{galvanometer}} = V_{\text{shunt}} \]
\[ I_g \times R_g = (I - I_g) \times S \]

Step 3: Substitute the known values and solve the algebra
Insert the meticulously collected numerical values directly into our established equation:
\[ I_g \times 27 = (10 - I_g) \times 3 \]
We can cleanly simplify the math by immediately dividing both sides entirely by 3:
\[ I_g \times 9 = 10 - I_g \]
Carefully rearrange the resulting terms to properly group the unknown variable $I_g$ onto the left side:
\[ 9 I_g + I_g = 10 \]
Combine the grouped terms:
\[ 10 I_g = 10 \]
Perform the final basic division to isolate $I_g$:
\[ I_g = \frac{10}{10} = 1 \text{ mA} \]

Step 4: Conclusion
The galvanometer coil itself requires precisely 1 mA of circulating current to safely reach its extreme full-scale deflection point. This elegantly and perfectly matches option (C).