In the circuit shown, when switch `S_(1)` is closed and `S_(2)` is open, the ideal voltmeter shows a reaiding of `18 V`. When switch `S_(2)` is closed and `S_(1)` is open, the reading of voltmeter is `24 V`. When `S_(1)` and `S_(2)` both are closed, the voltmeter reading will be

Options

A. `14.4 V`
B. `20.6 V`
C. `24.4 V`
D. `10.8 V`

Answer

A

Explanation

(a) Terminal potential difference across the terminals of cell `E`
`V = E - ir = E - ((E)/(r + R)) r = (ER)/(R + r)`
According to question
`18 = (6 E)/(6 + r)`
and `24 = (12 E)/(12 + r)`
Solving (i) and (ii), we get `r = 6 Omega, E = 36 V`
When both `S_(1)` and `S_(2)` are closed, the effective external resistance in the circuit will be.
`R = (6 xx 12)/(18) = 4 Omega`
So, `V = (ER)/(r + R) = (36 xx 4)/(6 + 4) = (144)/(10) = 144` volt
`V = 14.4` volt
Thus, choice (a) is correct.