Question

A moving coil instrument gives a full scale deflection of 20mA when potential difference across its terminal is 200mV. Calculate (i) the shunt resistance for measuring current up to 20A (ii) the series resistance for measuring voltage up to 400V.

Answer 1

(i) The current flowing through the shut,

$I_{x}=I-I_{m}$=(20×20×$10^{-3}$)= 19.98 A

$R_{x}=\frac{200*10^{-3} }{19.98}= 0.01001 ohms$

(ii) Total voltage = 400V

Voltage across the terminal = 200 mV

Full scale deflection = 20mA

The same current will flow through the series resistance also. The voltage drop across the resistance should be $(400-200*10^{-3})$ volts.

Answer 2

Answer:

shunt resistance for measuring current up to 20A is 19.98amps

the series resistance for measuring voltage up to 400V is 0.01001ohms

Explanation:

Given,

Current flowing through the shunt,

=total current through the shunt resistance-current through the moving coil

=

$20 - (20 \times {10}^{ - 3} )$

$19.98amps$

Voltage drop across the instrument terminal A and B shunt is 200m V. Same voltage drop will be across the shunt resistance also. Therefore

$IR = 200mV \\ = 200× {10}^{ - 3} V$

$Rs = \frac{2 00\times {10}^{ - 3} }{19.98}$

$0.01001ohms$

Therefore

series resistance for measuring voltage up to 400V is 0.01001ohms