A cell of e.m.f 2V and internal resistance 1(Omega) supplies a current of 0.5 amps to a resistance whose value is
-
A.
0.5(Omega) -
B.
1(Omega) -
C.
2(Omega) -
D.
2.5(Omega) -
E.
3(Omega)
Correct Answer: Option E
Explanation
1 = (frac{E}{R + r})
(frac{2}{R + 1}) = 0.5
0.5R = 2 – 0.5 = 1.5
R = (frac{1.5}{0.5})
= 3(Omega)