Question:
A d’Arsonval meter with an internal resistance of 1 kΩ requires 10 mA to produce full-scale deflection. Calculate the value of a series resistance needed to measure 50 V of full scale.
Answer:
4kΩ
Step-by-step explanation:
Given;
internal resistance, r = 1kΩ
current, I = 10mA = 0.01A
Voltage of full scale, V = 50V
Since there is full scale voltage of 50V, then the combined or total resistance (R) of the circuit is given as follows;
From Ohm's law
V = IR
R =
[substitute the values of V and I]
R =

R = 5000Ω = 5kΩ
The combined resistance (R) is actually the total resistance of the series arrangement of the series resistance(
) and the internal resistance (r) in the circuit. i.e
R =
+ r
= R - r [Substitute the values of R and r]
= 5kΩ - 1kΩ
= 4kΩ
Therefore the series resistance is 4kΩ