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Physics-

General

Easy

Question

A 5V battery with internal resistance 2 $blank capital omega$ and a 2V battery with internal resistance 1 $blank capital omega$ are connected to a 10 $blank capital omega$ resistor as shown in the figure

The current in the 10 $blank capital omega$ resistor is

0.27A, $P_{2} t o P_{1}$
0.03A, $P_{1} t o P_{2}$
0.03A, $P_{2} t p P_{1}$
0.27A, $P_{1} t o P_{2}$

The correct answer is: 0.03A, $P subscript 2 end subscript t p P subscript 1 end subscript$

Let potential of $P subscript 1 end subscript$ is 0 V and potential of $P subscript 2 end subscript$ is $V subscript 0 end subscript$ . Now apply KCL at $P subscript 2 end subscript$ .

$fraction numerator V subscript 0 end subscript minus 5 over denominator 2 end fraction plus fraction numerator V subscript 0 end subscript minus 0 over denominator 10 end fraction plus fraction numerator V subscript 0 end subscript minus open parentheses negative 2 close parentheses over denominator 1 end fraction equals 0$
$⟹ V subscript 0 end subscript equals fraction numerator 5 over denominator 16 end fraction$
So, current through 10 $capital omega$ resistor is
$fraction numerator V subscript 0 end subscript over denominator 10 end fraction$ from $P subscript 2 end subscript t o P subscript 1 end subscript$ .

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