Answer to Question #81536 in Electric Circuits for Ranjan

Question

A voltmeter coil has resistance 50 ohm and a resistor of 1.15k ohm is connected in series. It can read potential difference upto 12volts . If the same coil is used to construct an ammeter which can measure current upto 2A, what should be the resistance of the shunt used?

Answer 1

Coil has resistance:
R_coil=50Ohm
Resistor resistance:
R=1150Ohm
Read potential:
U=12V
Current:
I=2A
Solution:
Current flowing through the coil:
I_coil=U/(R+R_coil )=12/1200=0.01A
this current deflects the full-scale arrow and we have the voltage drop across the coil:
U_coil=I_coil R_coil=0.01*50=0.5V
This means that the current passing through the shunt will be equal to:
I_shunt=I-I_coil=2-0.01=1.99A
When connected in parallel, the voltage on the coil and shunt are the same:
U_shunt=U_coil
Hence the resistance of the shunt:
R_shunt=U_shunt/I_shunt =0.5/1.99=0.251Ohm