A galvanometer coil having a resistance of 20 Ω and a full-scale deflection at 1.0 mA is connected in series with a 4980 Ω resistance to build a voltmeter. What is the maximum voltage that this voltmeter can read?

E = IR + Ir

Where,

I = current

= 1 mA

= 0.001 A

R = 4980 Ω

Internal resistance, r = 20 Ω

Maximum voltage, E = 0.001 * 4980 + 0.001 * 20

= 4.98 + 0.02

= 5 V

5 V

Explanation:

The maximum voltage the voltmeter can read will be the voltage drop across the 20 Ω resistance and the voltage drop across the 4980 Ω resistance.

V' = Ir + IR ... equation 1

Where V' = Maximum voltage the voltmeter can read, I = current, r = resistance of the galvanometer coil, R = The resistance connected in series to the galvanometer.

Given: I = 1 mA = 0.001 A, r = 20Ω, R = 4980Ω

Substitute into equation 1

V' = 20 (0.001) + 4980 (0.001)

V' = 0.02+4.98

V' = 5 V