Find the current in the 6,Omega resistance. | vedclass

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(C) The circuit consists of a $10\,V$ battery in series with a $3\,\Omega$ resistor,which is in parallel with a branch containing $2\,\Omega$ and $1\,

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

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(C) The circuit consists of a $10\,V$ battery in series with a $3\,\Omega$ resistor,which is in parallel with a branch containing $2\,\Omega$ and $1\,\Omega$ resistors in series. This entire combination is in series with a $6\,\Omega$ resistor.First,calculate the equivalent resistance of the parallel part:$R_p = (2 + 1) \parallel 3 = \frac{3 	imes 3}{3 + 3} = \frac{9}{6} = 1.5\,\Omega$.Now,the total resistance of the circuit is:$R_{eq} = R_p + 6\,\Omega = 1.5 + 6 = 7.5\,\Omega$.The total current $I$ flowing from the battery is:$I = \frac{V}{R_{eq}} = \frac{10}{7.5} = \frac{100}{75} = \frac{4}{3}\,A$.Since the $6\,\Omega$ resistor is in series with the entire parallel combination,the total current $I$ flows through it.Therefore,the current in the $6\,\Omega$ resistance is $\frac{4}{3}\,A$.

Find the current in the $6\,\Omega$ resistance.

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