A galvanometer has full scale deflection current of 500 micro-Ampere, and resistance of 100 Ohm. Shunt with what resistance has to be combined with the galvanometer in order to transform it to an ammeter able to measure current of 10 Ampere full scale?

Rs = (RgIg) / (I - Ig) = (100 500 * 10^-6) / (10 - 50010^-6) = 0.005 ohm

Ig = 50010^-6 A

Rg = resistance of galvanometer = 100 ohm

I = 10 A

Student uses the circuit shown in task [6] in the text to find the resistance of a voltmeter, and obtains that at 2 Volt reading on the voltmeter, the galvanometer reads current of 400 micro-Ampere. What is the voltmeter resistance?

R = (V/Ig) - G = (2/40010^-6) - 40010^-6 = 5000 ohm

V = 2 V

Ig = 400 * 10^-6 A

Question

Alijah Townsend

6 August, 00:59

+4

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Nina Mcgee · Answer 1

the solution is part of it already