A resistance R is connected across the terminal of an electric cell of internal resistance 2Ω and the voltage was reduced to 3/5 of its nominal value.
The value of R is
-
3 Ω
- 2 Ω
- 1 Ω
- 6 Ω
The correct answer is: A
Explanation
I = \(\frac{V}{R}\) = \(\frac{E}{R+r}\) , if r = 2Ω
V = \(\frac{3}{5}\)E
=> \(\frac{3E}{5R} = \frac{E}{R+2}\)
i.e 3E(R+2) = 5RE
3RE + 6E = 5RE ( the Es will cancel out)
i.e 3R + 6 = 5R
6 = 5R - 3R
=> 2R = 6 i.e R = \(\frac{6}{2}\) = 3Ω
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