A 5 V battery with internal resistance 2 - and a 2 V battery with internal resistance 1 - are connected to a 10 - resistor as shown in figure-The current in the 10 - resistor isA- 0-03 A - P 1 to P 2B- 0-03 A - P 2 to P 1C- 0-27 A - P 1 to P 2D- 0-27 A - P 2 to P 1

The correct option is B 0.03A, P2 to P1 ∴ Current throught 10Ω resister =l_1 l_2 =1/32A Loop(1) =0.03A from P_2 to P_1 12l_1 10l_2=5 ——— ...

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Byju's Answer

Standard XII

Physics

Parallel Plate Capacitor

A 5 V battery...

Question

A 5V battery with internal resistance 2 $Ω$ and a 2V battery with internal resistance 1 $Ω$ are connected to a 10 $Ω$ resistor as shown in figure.

The current in the 10 $Ω$ resistor is

0.27A, P₂ to P₁

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0.03A, P₁ to P₂

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0.03A, P₂ to P₁

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0.27A, P₁ to P₂

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Solution

The correct option is C
0.03A, P₂ to P₁

$∴$ Current throught 10 $Ω$ resister
$= l_{1} - l_{2}$
$= \frac{1}{32} A$ Loop(1)
=0.03A from $P_{2}$ to $P_{1}$ $12 l_{1} - 10 l_{2} = 5$ ---------(1)
Loop(2)
$- 10 l_{1} + 11 l_{2} = 2$ ----------------(2)
Solving (1) & (2)
$l_{2} = 75 / 32 A$
$I_{1} = 74.5 / 32 A$

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A 5V battery with internal resistance 2Ω and a 2V battery with internal resistance 1Ω are connected to a 10Ω resistor as shown in figure. The current in the 10Ω resistor is

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A 5V battery with internal resistance 2 $Ω$ and a 2V battery with internal resistance 1 $Ω$ are connected to a 10 $Ω$ resistor as shown in figure.

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