Question

A galvanometer having a resistance of $50 \Omega$, gives a full scale deflection for a current of 0.05 A. The length in meters of a resistance wire of area of cross section 3 × 10-2 cm2 that can be used to convert the galvanometer into an ammeter which can read a maximum of 5 A current is

(Specific resistance of the wire $\rho = 5 \times 10 ^ { - 7 } \Omega \mathrm { m }$ )

• A. 9
• B. 6
• C. 3
• D. 1.5

Answer 1

Answer: (C)

3

Answer 2

$\mathrm { S } = \frac { \mathrm { I } _ { \mathrm { g } } \mathrm { G } } { \mathrm { I } - \mathrm { I } _ { \mathrm { g } } } = \frac { 0.05 \times 50 } { 5 - 0.5 } = \frac { 2.5 } { 4.95 } = \frac { 250 } { 495 } = \frac { 50 } { 99 } \Omega$

$\because S = \frac { \rho l } { A }$ or $1 = \frac { S A } { \rho }$

$\therefore 1 = \frac { 50 } { 99 } \times \frac { \left( 3 \times 10 ^ { - 6 } \mathrm {~m} ^ { 2 } \right) } { \left( 5 \times 10 ^ { - 7 } \Omega \mathrm {~m} \right) } = 3.0 \mathrm {~m}$