Two batteries one of emf 18 volts and internal resistance 2 - and the other of emf 12 V and internal resistance 1 - are connected as shown in figure the voltmeter V will record a reading ofA- 18 VB- 30 VC- 14 VD- 15 V

The correct option is C 14VI=E_2 E_1/r_2+r_1=18 12/2+1=6/3 I=2A V_1 12 2× 1=V_2 V = V_1 V 2=12+2=14V ...

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Byju's Answer

Standard XII

Physics

Non-Ideal Battery

Two batteries...

Question

Two batteries one of emf 18 volts and internal resistance 2 $Ω$ and the other of emf 12V and internal resistance 1 $Ω$ are connected as shown in figure the voltmeter V will record a reading of

18V

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30V

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14V

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15V

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Solution

The correct option is C 14V
$I = \frac{E_{2} - E_{1}}{r_{2} + r_{1}} = \frac{18 - 12}{2 + 1} = \frac{6}{3} I = 2 A V_{1} - 12 - 2 \times 1 = V_{2} V = V_{1} - V - 2 = 12 + 2 = 14 V$

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Two batteries one of emf 18 volts and internal resistance 2 Ω and the other of emf 12V and internal resistance 1 Ω are connected as shown in figure the voltmeter V will record a reading of

The correct option is C 14VI=E2−E1r2+r1=18−122+1=63I=2AV1−12−2×1=V2V=V1−V−2=12+2=14V

Two batteries one of emf 18 volts and internal resistance 2 $Ω$ and the other of emf 12V and internal resistance 1 $Ω$ are connected as shown in figure the voltmeter V will record a reading of

The correct option is C 14V
$I = \frac{E_{2} - E_{1}}{r_{2} + r_{1}} = \frac{18 - 12}{2 + 1} = \frac{6}{3} I = 2 A V_{1} - 12 - 2 \times 1 = V_{2} V = V_{1} - V - 2 = 12 + 2 = 14 V$