Question

When moving coil galvanometer (MCG) is converted into a voltmeter, the series resistance is '$n$' times the resistance of galvanometer. How many times that of MCG the voltmeter is now capable of measuring voltage?

• A. $n$
• B. $\frac{n+1}{n}$
• C. $n + 1$
• D. $n - 1$

Hint:

The voltage multiplication factor $m$ is always $1 + \frac{R_s}{G}$. If $R_s = nG$, then $m = n + 1$.

Answer 1

The Correct Option is C

Step 1: Let $G$ be the resistance of the galvanometer and $I_g$ be the current required for full-scale deflection. The maximum voltage it can measure as a galvanometer is: \[ V_g = I_g G \]
Step 2: To convert the galvanometer into a voltmeter, a high resistance $R_s$ is connected in series. The new voltage range $V$ is given by: \[ V = I_g (G + R_s) \]
Step 3: We are given that the series resistance is $n$ times the galvanometer resistance, so $R_s = nG$. Substituting this into the formula for $V$: \[ V = I_g (G + nG) \]
Step 4: Simplify the expression: \[ V = I_g G (1 + n) \]
Step 5: Since $V_g = I_g G$, we can write: \[ V = V_g (n + 1) \] Thus, the new voltmeter is capable of measuring $n + 1$ times the initial voltage.