Two cells each of e.m.f. 1.5V and an internal resistance 2(Omega) are connected in parallel. Calculate the current flowing when the cells are connected to a 1(Omega) resistor
-
A.
0.75(Omega) -
B.
1.5(Omega) -
C.
0.5(Omega) -
D.
1.0(Omega) -
E.
0.6(Omega)
Correct Answer: Option A
Explanation
E = 1.5v
rT = (frac{2 times 2}{2 + 2})
= 1
e.m.f. = I(R + r)
1.5 = I(1 + 1)
I = 0.75(Omega)