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When an ammeter of negligible internal resistance is inserted in series with circuit, it reads 1 A. When the voltmeter of very large resistance is connected across R1, it reads 3 V. But when the points A and B are short-circuited by a conducting wire, then the voltmeter measures 10.5 V across the battery. The internal resistance of the battery is equal to

Option: 1

\mathrm{\frac{3}{7}}\Omega


Option: 2

5 \Omega


Option: 3

3 \Omega


Option: 4

None of These


Answers (1)

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\mathrm{3=I R_1~ or~ 3=1 \times R_1 or R_1=3 \Omega }When points

A and B are connected by a conducting wire, \mathrm{R_2} is short-circuited.

\begin{aligned} & \mathrm{\therefore 10.5=I^{\prime} R_1 \text { or } 10.5=I^{\prime} \times 3 }\\ &\mathrm{ \therefore \quad I^{\prime}=\frac{10.5}{3}=3.5 \mathrm{~A} }\end{aligned}
But \mathrm{10.5=E-I^{\prime} r or 10.5=12-3.5 r }

\mathrm{\therefore r=\frac{1.5}{3.5}=\frac{3}{7} \Omega }

Posted by

Ritika Jonwal

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