Question 35: Current Electricity

Question

A galvanometer of $50 \Omega$ resistance is converted into an ammeter using a shunt resistance of $10 \Omega$. If the same resistance is used to convert the same galvanometer in to a voltmeter, what would be the ratio of the resistance of the ammeter to the resistance of the voltmeter?

Accepted Answer

Let G=50 ÃÂ© and the given resistor S=10 ÃÂ©.
Ammeter resistance
When you put S in parallel with G,
$$R_A=\frac{G\cdot S}{G+S}=\frac{50	imes10}{50+10}=\frac{500}{60}=\frac{25}{3}\,\Omega.$$
Voltmeter resistance
When you use the same S in series with G,
$$R_V=G+S=50+10=60\,\Omega.$$
Ratio
$$R_A:R_V=\frac{25/3}{60}=\frac{25}{180}=\frac{5}{36}.$$
Answer: 5 : 36 (Option B).

Answer

$5: 1$

Answer

$5: 36$

Answer

$5: 6$

Answer

$1: 5$

Current Electricity

A galvanometer of 50 \Omega resistance is converted into an ammeter using a shunt resistance of 10 \Omega. If the same resistance is used to convert the same galvanometer in to a voltmeter, what would be the ratio of the resistance of the ammeter to the resistance of the voltmeter?

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