A 60 microampere meter has a resistance of 100 ohms. If the meter has to measure 100 mA, then the value of current through the shunt is

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ALP CBT 2 Electrician Previous Paper: Held on 23 Jan 2019 Shift 2

Answer 1

Concept:

\({R_{sh}} = \frac{{{R_m}}}{{M - 1\;}}\)

Where,

R_sh= Shunt resistance

R_m= meter resistance

M = Mutiplying factor

M = (Present value)/(Pre-existing value)

Calculation:

\(M = \;\frac{{100 \times {{10}^{ - 3}}}}{{60 \times {{10}^{ - 6}}}} = 1666.67\;\)

\({R_{sh}} = \frac{{100}}{{1666.67 - 1\;}} = \frac{{100}}{{1665.67}} = 0.06\;{\rm{\Omega }}\)

Now by applying current division rule in the 2^nd circuit, we get

\({I_{sh}} = \frac{{100 \times {{10}^{ - 3}} \times 100}}{{100.06}} = 99.9\;mA\)