Question

A galvanometer of resistance G has voltage range Vg. Resistance required to convert it to read voltage up to V is

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

Answer 1

Answer: (B)

G($\frac {V}{V_g}$-1)

Answer 2

The total resistance in the circuit is given by the sum of the galvanometer resistance and the series resistor: G + R.
Using Ohm's Law, the current passing through the galvanometer can be calculated as I =$\frac { V}{G + R}$
Since we want to limit the current to be within the galvanometer's range, we have: I ≤ $\frac { V_g}{G}$.
Substituting the expression for I and rearranging the inequality, we get:
$\frac { V}{G + R}$ ≤ $\frac { V_g}{G}$
Multiplying both sides by G and rearranging, we have:
V ≤ $(\frac { V_g}{G})$ * (G + R)
V ≤ Vg + $(\frac { V_g}{G})$ * R
Now, subtracting Vg from both sides, we get:
V - Vg ≤ $(\frac { V_g}{G})$ * R
Dividing both sides by $(\frac { V_g}{G})$, we have:
$\frac { (V - Vg)}{(Vg / G)}$ ≤ R
R ≥ (V - Vg) * $\frac {G}{V_g}$
R ≥$\frac { VG - V_gG}{V_g }$
R ≥ $\frac {V*G}{V_g}$ - G
R ≥ G*($\frac { V}{V_g}$- 1)
Therefore, the correct option is (B) G($\frac { V}{V_g}$ - 1), which represents the resistance required to convert the galvanometer to read voltage up to V.