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Standard XII

Physics

Kirchhoff's Junction Law

The currents ...

Question

The currents $I_{1}, I_{2}, & I_{3}$ in the circuit are respectively:

1/2, 1/2, 0

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1, 1, 2

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2, 2, 1

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1, 2, 3

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Solution

The correct option is B 1/2, 1/2, 0
By using kirchoffs law we have
$\frac{x - 4.5}{3} + \frac{x - 3}{10} + \frac{x - 0}{6} = 0$
$\Rightarrow \frac{10 x - 45 + 3 x - 9 = 5 x}{30} = 0$
$\Rightarrow 18 x = 54$
$x = \frac{54}{18}$
$∴ i_{1} = \frac{1}{2}$ $i_{2} = \frac{1}{2}$ $i_{3} = 0$
Hence, option $A$ is correct answer.

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The currents I1,I2,&I3 in the circuit are respectively:

The correct option is B 1/2, 1/2, 0By using kirchoffs law we havex−4.53+x−310+x−06=0⇒10x−45+3x−9=5x30=0⇒18x=54x=5418∴i1=12 i2=12 i3=0Hence, option A is correct answer.

The currents $I_{1}, I_{2}, & I_{3}$ in the circuit are respectively: