Figure (32 - E10) shows a part of an electric circuit. The potentials at teh points a,b and c are 30 V , 12 V and 2 V respectively. Find the currents through the three resistors.

Figure (32 - E10)

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Solution

Let potential at the point be $x$ V

Then (30 - x) = 10 $i_{1}$

( $x - 12$ ) = 20 $i_{2}$

( $x - 2$ ) = 30 $i_{3}$

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

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

$\Rightarrow 30 - x = \frac{x - 12}{2} + \frac{x - 2}{3}$

$\Rightarrow 30 - x = \frac{3 x - 36 + 2 x - 4}{6}$

$\Rightarrow 180 - 6 x = 5 x - 40$

$\Rightarrow 11 x = 220$

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

$i_{1} = \frac{30 - 20}{10} = 1 A$

$i_{2} = \frac{20 - 12}{20} = \frac{8}{20}$

= $0.4 A$ $i_{3} = \frac{20 - 2}{30} A = \frac{18}{30} A$

= $\frac{6}{10} A = 0.6 A$

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