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Question

Physics Question on Current electricity

In the given circuit, an ideal voltmeter connected across the 10Ω10\, \Omega resistance reads 2V2V. The internal resistance rr, of each cell is :

A

1Ω1 \Omega

B

1.5Ω1.5 \Omega

C

0Ω0\Omega

D

0.5Ω0.5\Omega

Answer

0.5Ω0.5\Omega

Explanation

Solution

Req=15×1025+2+2rR_{eq} = \frac{15\times10}{25} + 2 +2r
=8+2r= 8 + 2r
i=38+2ri = \frac{3}{ 8+2r}
2=iReq=38+2r×62 = i R_{eq} = \frac{3}{8 + 2 r } \times6
16+4r=1816 +4r = 18
r=0.5Ω\Rightarrow r = 0.5 \Omega