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(C) The circuit consists of a $14\,V$ battery in series with a $5\,\Omega$ resistor,which is then connected in parallel to two branches: one with a $3

Vedclass · Accepted Answer

$\frac{4}{3}\,A$

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(C) The circuit consists of a $14\,V$ battery in series with a $5\,\Omega$ resistor,which is then connected in parallel to two branches: one with a $3\,\Omega$ resistor and another with a $(2+4)\,\Omega = 6\,\Omega$ resistor.First,calculate the equivalent resistance of the parallel part:$R_p = \frac{3 	imes 6}{3 + 6} = \frac{18}{9} = 2\,\Omega$The total equivalent resistance of the circuit is:$R_{eq} = R_p + 5\,\Omega = 2\,\Omega + 5\,\Omega = 7\,\Omega$The total current $I$ flowing from the battery is:$I = \frac{V}{R_{eq}} = \frac{14\,V}{7\,\Omega} = 2\,A$Using the current divider rule,the current $I_1$ through the $3\,\Omega$ resistor is:$I_1 = I 	imes \left( \frac{R_{other}}{R_{3\Omega} + R_{other}} ight) = 2\,A 	imes \left( \frac{6\,\Omega}{3\,\Omega + 6\,\Omega} ight) = 2 	imes \frac{6}{9} = 2 	imes \frac{2}{3} = \frac{4}{3}\,A$

Find the current in the $3\,\Omega$ resistance in the given circuit.

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