Question 26: Current Electricity

Question

A moving coil galvanometer has a resistance of 60 $$\Omega$$ and it indicates full deflection on passing a current of 4.5 mA. A voltmeter is made using this galvanometer and a 4.5 k$$\Omega$$ resistance. The maximum voltage, that can be measured using this voltmeter, will be close to

Accepted Answer

Given, resistance of galvanometer, R
g
= 60$$\Omega$$
Current, I
g
= 4.5 mA = 4.5 $$	imes$$ 10
$$-$$3
A
Resistance used in converting a galvanometer into voltmeter, R = 4.5 k$$\Omega$$ = 4.5 $$	imes$$ 10
3
$$\Omega$$
To make the voltmeter, the galvanometer and the high resistance (R) have to be connected in series combination as shown below.
Image
$$	herefore$$ Maximum current in galvanometer is
$${I_g} = {E \over {R + {R_g}}}$$
$$ \Rightarrow E = {I_g}(R + {R_g})$$
$$ = 4.5 	imes {10^{ - 3}} 	imes (4.5 	imes {10^3} + 60)$$
$$ = 20.5$$ V

Answer

21 V

Answer

20.5 V

Answer

20 V

Answer

19.5 V

Current Electricity

A moving coil galvanometer has a resistance of 60 $\Omega and it indicates full deflection on passing a current of 4.5 mA. A voltmeter is made using this galvanometer and a 4.5 k\Omega$ resistance. The maximum voltage, that can be measured using this voltmeter, will be close to

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