Question

A galvanometer has resistance G and range Vg. How much resistance is required to read voltage up to V volt?

• A. \(GV_g\)
• B. \(G(\frac{V-V_g}{V})\)
• C. \(G(\frac{V}{V_g}-1)\)
• D. \(G(\frac{V+V_g}{V})\)

Hint:

The total resistance must give full-scale current at voltage V.

Answer 1

The Correct Option is C

Step 1: Voltage division
Voltage across the galvanometer must be $V_g$ when the total is $V$. Same current flows through both: $\frac{V_g}{G} = \frac{V-V_g}{R}$.

Step 2: Solve
$R = G\frac{V-V_g}{V_g} = G\left(\frac V{V_g}-1\right)$. Option (C).

Final Answer:
G(V/Vg - 1). \[ \boxed{\text{(C)}\ G\left(\frac{V}{V_g}-1\right)} \]