Figure shows a part of an electric circuit. The potentials at points a, b and c are 30 V, 12 V and 2 V respectively. The current through the 20 resistor will be. [2023]

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Solution

Q.
Figure shows a part of an electric circuit. The potentials at points a, b and c are 30 V, 12 V and 2 V respectively. The current through the 20 $Ω$ resistor will be. [2023]

1 0.4 A

2 0.2 A

3 0.6 A

4 1.0 A

Ans.
(1)

Sum of current at junction point will be zero:

$\frac{x - 30}{10} + \frac{x - 12}{20} + \frac{x - 2}{30} = 0$

$\Rightarrow x (\frac{1}{10} + \frac{1}{20} + \frac{1}{30}) = \frac{30}{10} + \frac{12}{20} + \frac{2}{30}$

$\Rightarrow x (\frac{6 + 3 + 2}{60}) = \frac{180 + 36 + 4}{60}$

$\Rightarrow x = \frac{220}{11} = 20 V$

$∴$ $Current through 20 Ω, I_{20 Ω} = \frac{x - 12}{20}$

$\Rightarrow I_{20 Ω} = \frac{20 - 12}{20} = \frac{2}{5} = 0.4 A$

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