Find the current flowing through the resistance R1 of the circuit shown in figure- if the resistances are equal to R1-10 - R2-10 - and R3-10 - and the potential of points 1- 2 and 3 are equal to V 1-10 V- V 2-6 V and V 3-5 V-

Assume potential at the junction to be x and apply KCL. i_1 + i_2 + i_3 = 0 ⇒x 1010 + x 610 + x 510 = 0 ⇒ x = 7 V So, i_1 = x 1010 =7 1010= 0.3 A So, the ...

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Byju's Answer

Standard XII

Physics

Series and Parallel Combination of Resistors

Find the curr...

Question

Find the current flowing through the resistance $R_{1}$ of the circuit shown in figure, if the resistances are equal to $R_{1} = 10 Ω$ , $R_{2} = 10 Ω$ and $R_{3} = 10 Ω$ , and the potential of points $1, 2$ and $3$ are equal to $V_{1} = 10 V$ , $V_{2} = 6 V$ and $V_{3} = 5 V$ .

0.1 A

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0.2 A

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0.3 A

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0.4 A

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Solution

The correct option is C $0.3 A$

Assume potential at the junction to be $x$ and apply KCL.

$i_{1} + i_{2} + i_{3} = 0$

$\Rightarrow \frac{x - 10}{10} + \frac{x - 6}{10} + \frac{x - 5}{10} = 0$

$\Rightarrow x = 7 V$

So,

$i_{1} = \frac{x - 10}{10} = \frac{7 - 10}{10} = - 0.3 A$

So, the direction of current will be opposite to the assumption.

Therefore, option $(c)$ is correct.

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