Numerical explaning Shunt resistance for Extending the Range of PMMC as an Ammeter (05:24)

Lecture Notes :-

Let us see the numerical explaining shunt resistance for extending the range of PMMC as an Ammeter.

Q. A 100 μA ammeter has an internal resistance of 100Ω. For extending its range to measure 500 μA, the shunt resistance required is of
[gate – 2001]

a) 20 Ω

b) 22.22 Ω

c) 25 Ω

d) 50 Ω

Ans. Given that, R_m = 100 Ω

I_m = 100 μA

I = 500 μA

R_sh = ?

We know the formula of shunt resistance, it is given by,

R_{sh}=\frac{R_{m}}{m-1}

The multiplying factor is given by,

m=\frac{I}{I_{m}}

Putting values we get,

m=\frac{500\times 10^{-6}}{100\times 10^{-6}}=5

As R_{sh}=\frac{R_{m}}{m-1}

R_{sh}=\frac{100}{5-1}

R_sh = 25 Ω

So, option (c) is the correct answer.