Answer:
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
Answer:
Explanation:
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
