In the circuit shown in figure when switch S1 is closed and S2 is open- the ideal voltmeter shows a reading of 18V- When switch S2 is closed and S1 is open- the reading of the voltmeter id 24V- When S1 and S2 both are closed the voltmeter reading will be

V=E=i r=E (ER+r) r=E RR+r nbsp; Hence, 18=6 E/6+r⋯(i) And 24=12 E12+r⋯(ii) Solving Eqs. (i) and (ii), we get r=6 Ω and E=36 V When both S 1 and S ...

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Byju's Answer

Standard XII

Physics

Mutual Induction

In the circui...

Question

In the circuit shown in figure when switch S₁ is closed and S₂ is open, the ideal voltmeter shows a reading of 18V. When switch S₂ is closed and S₁ is open, the reading of the voltmeter id 24V. When S₁ and S₂ both are closed the voltmeter reading will be

14.4 V

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20.6 V

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24.2 V

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10.8 V

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Solution

The correct option is A
14.4 V

$V = E = i r = E - (\frac{E}{R + r}) r = \frac{E R}{R + r}$
$Hence, 18 = \frac{6 E}{6 + r} \dots (i)$

And $24 = \frac{12 E}{12 + r} \dots (i i)$

Solving Eqs. $(i)$ and $(i i),$ we get

$r = 6 Ω and E = 36 V$

When both $S 1$ and $S 2$ are closed external resistance in the circuit will be

$R = \frac{(6) (12)}{6 + 12} = 4 Ω$
$∴ V = \frac{(36) (4)}{4 + 6} = 14.4 V$

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In the circuit shown in figure when switch S1 is closed and S2 is open, the ideal voltmeter shows a reading of 18V. When switch S2 is closed and S1 is open, the reading of the voltmeter id 24V. When S1 and S2 both are closed the voltmeter reading will be

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