A galvanometer of resistance 50Ω is connected to a battery of 3 V along with a resistance

A galvanometer of resistance 50Ω is connected to a battery of 3 V along with a resistance of 2950 Ω in series. A full scale deflection of 30 divisions is obtained in the galvanometer. In order to reduce this deflection to 20 divisions, the resistance in series should be

Options

(a) 6050 Ω
(b) 4450 Ω
(c) 5050 Ω
(d) 5550 Ω

Correct Answer:

4450 Ω

Explanation:

I=3 / (50+2950) = 10⁻³ A

Current for 30 divisions=10⁻³ A

Current for 20 divisions=10⁻³ * 20/30

=2*10⁻³ / 3 Amperes

For the same deflection to obtain for 20 divisions let resistance added be R, therefore,

2*10⁻³ / 3 = 3 / (50+R)

(50+R) = 9 *10⁻³ / 2

(50+R) = 4500

Therefore R=4450Ω

admin:

Related Questions

  1. Two circuits have a mutual inductance of 0.1 H. What average e.m.f. is induced
  2. A dip needle vibrates in a vertical plane perpendicular to magnetic meridian. The time
  3. The fluctuation in the input voltage of 200 V to a domestic circuit is +10V
  4. A source of sound S emitting waves of frequency 100 Hz and an observer
  5. Small drops of the same size are charges to V volt each. If n such drops coalesce