Magnetics 6 Question 5
5. A moving coil galvanometer has resistance $50 \Omega$ and it indicates full deflection at $4 mA$ current. A voltmeter is made using this galvanometer and a $5 k \Omega$ resistance. The maximum voltage, that can be measured using this voltmeter, will be close to
(2019 Main, 9 April I)
(a) $40 V$
(b) $10 V$
(c) $15 V$
(d) $20 V$
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Solution:
- Given, resistance of galvanometer, $R _g=50 \Omega$
Current, $I _g=4 mA=4 \times 10^{-3} A$
Resistance used in converting a galvanometer in voltmeter, $R=5 k \Omega=5 \times 10^{3} \Omega$
$\therefore$ Maximum current in galvanometer is
$$ I _g=\frac{E}{R+R _g} $$
$$ \begin{aligned} \therefore \quad E & =I _g\left(R+R _g\right) \\ & =4 \times 10^{-3} \times\left(5 \times 10^{3}+50\right) \\ & =5050 \times 4 \times 10^{-3} \\ & =20.2 V \simeq 20 V \end{aligned} $$
