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

Answer

`14.4 V`

Explanation

As `V = epsilon - Ir and I = (epsilon)/(R + r)`
`:. V = IR = (epsilon R)/(R + r) or 18 = (epsilon xx 6)/(6 + r)`…..(i)
and `2A = (epsilon xx 12)/(12 + r)`….(ii)
Solving (i) and (ii) we get
`r = 6 Omega and epsilon = 36 V`
when both the swither `S_(1) and S_(2)` are closed then effective external resistance is the circuit will be
`R = (6 xx 12)/( 6+ 12) = 4 Omega`
`:. V = (epsilon R )/(R + r) = (36 xx 4)/( 4+ 6) = 14.4 V`