A capacitor of `4 mu F` is connected as shown in the circuit. The internal resistance of the battery is `0.5 Omega`. The amount of charge on the capacitor plates will be
.

Options

A. 0
B. `4 mu C`
C. `16 mu C`
D. `8 mu C`

Answer

D

Explanation

In Fig, no current si allowed to flow through `10 Omega` resistance because of capacitor. The current drawn by `2 Omega` res.
`I = (E)/(R + r) = (2.5)/(2 + 0.5) =1 A`.
Terminal pot. Diff of battery, `V = E - Ir = 2.5 - 1xx0.5 = 2` volt
`:.` Charge on capacitor plates, `Q = CV = 4xx2 = 8 muC`
Choice (d) is correct.