Question 512: Current Electricity

Question

The scale of a galvanometer is divided into 160 equal divisions. The galvanometer shows full scale deflection of 16 mA and maximum voltage is 80 mV . Now the range is changed so that galvanometer reads 160 V . The required resistance to be connected is

Accepted Answer

Given: $\mathrm{I}_{\mathrm{g}}=16 \mathrm{~mA}=16 	imes 10^{-3} \Omega$
$$\mathrm{V}_{\mathrm{g}}=80 \mathrm{~mA}=80 	imes 10^{-3} \Omega$$
The galvanometer's internal resistance $R_g$ will be:
$$\mathrm{R}_{\mathrm{g}}=\frac{\mathrm{V}_{\mathrm{g}}}{\mathrm{I}_{\mathrm{g}}}=\frac{80 	imes 10^{-3}}{16 	imes 10^{-3}}=5 \Omega$$
To increase the range of the galvanometer to 160 V , resistance needs to be connected in series to limit the current through the galvanometer to its full-scale deflection current when 160 V is applied.
This configuration converts the galvanometer into a voltmeter with a range of 160 V .
For a voltmeter, the total voltage (V) across the galvanometer and the series resistor is given by:
$$V_g=I_g\left(R_g+R_sight)$$
Substituting the values, we get:
$$\begin{aligned}
& 160=16 	imes 10^{-3}\left(5+R_sight) \
& R_s=\frac{160}{16 	imes 10^{-3}}-5=10000-5=9995 \Omega
\end{aligned}$$
Thus, the correct answer is $9995 \Omega$ in series.

Answer

$9995 \Omega$ in series.

Answer

$4995 \Omega$ in series.

Answer

$9.5 	imes 10^{-3} \Omega$ in parallel.

Answer

$4.95 	imes 10^{-3} \Omega$ in parallel.

Current Electricity

The scale of a galvanometer is divided into 160 equal divisions. The galvanometer shows full scale deflection of 16 mA and maximum voltage is 80 mV . Now the range is changed so that galvanometer reads 160 V . The required resistance to be connected is

Choose the correct answer: