Question

A galvanometer of resistance G has voltage range V_g. 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

The Correct Option is B

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.