Question

A galvanometer of resistance $G$ is converted into an ammeter using a shunt of resistance $R$. If the ratio of the heat dissipated through the galvanometer and shunt is $3:4$, then $R$ equals

• A. $\frac{4}{3}G$
• B. $\frac{3}{4}G$
• C. $\frac{16}{9}G$
• D. $\frac{9}{16}G$
• E. $G$

Hint:

For parallel circuits: same voltage, use $P = \frac{V^2}{R}$ when easier.

Answer 1

The Correct Option is D

Concept: Heat (or power) dissipated in a resistor: \[ P = I^2 R \] In a galvanometer–shunt combination:
• Both are connected in parallel
• Potential difference across both is same

Step 1: Let currents be \[ I_g = \text{current through galvanometer}, \quad I_s = \text{current through shunt} \]

Step 2: Given heat ratio \[ \frac{P_g}{P_s} = \frac{3}{4} \] Using $P = I^2 R$: \[ \frac{I_g^2 G}{I_s^2 R} = \frac{3}{4} \quad \cdots (1) \]

Step 3: Use same potential condition Since both are in parallel: \[ I_g G = I_s R \] \[ \Rightarrow \frac{I_g}{I_s} = \frac{R}{G} \quad \cdots (2) \]

Step 4: Substitute (2) into (1) \[ \frac{(R/G)^2 \cdot G}{R} = \frac{3}{4} \] \[ \Rightarrow \frac{R^2}{G^2} \cdot \frac{G}{R} = \frac{3}{4} \] \[ \Rightarrow \frac{R}{G} = \frac{3}{4} \]

Step 5: Careful reasoning This is current ratio relation — but power relation already includes square term. Using $P = \frac{V^2}{R}$ (better method since V same): \[ \frac{P_g}{P_s} = \frac{R}{G} \] \[ \frac{3}{4} = \frac{R}{G} \] But this is incorrect path (important correction). Correct approach: From $I_g G = I_s R$: \[ I_g = \frac{R}{G} I_s \] Substitute into power ratio: \[ \frac{(\frac{R}{G}I_s)^2 G}{I_s^2 R} = \frac{3}{4} \] \[ \frac{R^2}{G^2} \cdot \frac{G}{R} = \frac{3}{4} \] \[ \frac{R}{G} = \frac{3}{4} \] Final Answer: \[ R = \frac{3}{4}G \]