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  • Need clarity, kindly explain! - Current Electricity - JEE Main

A 5 V battery with internal resistance  2\Omega and 2 V battery with internal resistance 1\Omega  are connected to a 10\Omega  resistor as shown in the figure. The current in the10\Omega resistor is

 

  • Option 1)

    0.27 A\: \: P_{1}\: \: to\: \: P_{2}

  • Option 2)

    0.27 A\: \: P_{2}\: \: to\: \: P_{1}

  • Option 3)

    0.03 A\: \: P_{1}\: \: to\: \: P_{2}

  • Option 4)

    0.03 A\: \: P_{2}\: \: to\: \: P_{1}

 

Answers (1)

best_answer

As we learnt in

In closed loop -

-i_{1}}R_{1} + i_{2}}R_{2} -E_{1}-i_{3}}R_{3}+E_{2}+E_{3}-i_{4}}R_{4}=0

- wherein

 

 

Applying Kirchhoff’s law for the loops

AP_{2}P_{1}CA\: \: and \: \: P_{2}BDP_{1}P_{2}\: \: one \: \: gets

-10y-2x+5=0

\Rightarrow 2x+10y=5\cdots \cdots \cdots \cdots (i)

+2-1(x-y)+10.y=0

+x-11y=2\cdots \cdots \cdots \cdots (ii)

2x-22y= 4\cdots \cdots \cdots \cdots (iii)=(ii)\times 2

(i)-(iii)gives\ 32 y=1

\Rightarrow y=\frac{1}{32}A= 0.03A from P_{2}\: \: to\: \: P_{1}


Option 1)

0.27 A\: \: P_{1}\: \: to\: \: P_{2}

This option is incorrect.

Option 2)

0.27 A\: \: P_{2}\: \: to\: \: P_{1}

This option is incorrect.

Option 3)

0.03 A\: \: P_{1}\: \: to\: \: P_{2}

This option is incorrect.

Option 4)

0.03 A\: \: P_{2}\: \: to\: \: P_{1}

This option is correct.

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