In the circuit shown, the current through the `5 Omega` resistor is-

Options

A. `8/3 A`
B. `9/13 A`
C. `4/13 A`
D. `1/3 A`

Answer

D

Explanation

`E_(eq) = (E_(1)r_(2) +E_(2)r_(1))/(r_(1) + r_(2)) = 2V, r_(eq) = 2/2 = 1 Omega, I = 2/6 = 1/3 A`