Two batteries with e-m-f 12 V and 13 V are connected in parallel across a load resistor of 10 - The internal resistances of the two batteries are 1 - and 2 - respectivey- The voltage across the load lies between

From the junction law, we can write x 121 + nbsp;x 132 + nbsp;x 010 = 0 10x 120 + 5x 65 + x = 0 16 x = 185 x = 18516= 11.5625 Which lies in between 11. ...

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Standard XIII

Physics

EMF and EMF Devices

Two batteries...

Question

Two batteries with e.m.f $12 V$ and $13 V$ are connected in parallel across a load resistor of $10 Ω .$ The internal resistances of the two batteries are $1 Ω$ and $2 Ω$ respectivey. The voltage across the load lies between

11.7 V

and

11.8 V

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11.6 V

and

11.7 V

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11.5 V

and

11.6 V

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11.4 V

and

11.5 V

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Solution

The correct option is C $11.5 V$ and $11.6 V$

From the junction law, we can write $\frac{x - 12}{1} + \frac{x - 13}{2} + \frac{x - 0}{10} = 0$
$10 x - 120 + 5 x - 65 + x = 0$
$16 x = 185$
$x = \frac{185}{16} = 11.5625$
Which lies in between $11.5 V$ and $11.6 V$ .

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Two batteries with e.m.f 12 V and 13 V are connected in parallel across a load resistor of 10 Ω. The internal resistances of the two batteries are 1 Ω and 2 Ω respectivey. The voltage across the load lies between

The correct option is C 11.5 V and 11.6 V From the junction law, we can write x−121+x−132+x−010=0 10x−120+5x−65+x=0 16x=185 x=18516=11.5625 Which lies in between 11.5 V and 11.6 V.

Two batteries with e.m.f $12 V$ and $13 V$ are connected in parallel across a load resistor of $10 Ω .$ The internal resistances of the two batteries are $1 Ω$ and $2 Ω$ respectivey. The voltage across the load lies between

The correct option is C $11.5 V$ and $11.6 V$

From the junction law, we can write $\frac{x - 12}{1} + \frac{x - 13}{2} + \frac{x - 0}{10} = 0$
$10 x - 120 + 5 x - 65 + x = 0$
$16 x = 185$
$x = \frac{185}{16} = 11.5625$
Which lies in between $11.5 V$ and $11.6 V$ .