As shown in the figure,a network of resistors | vedclass

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(D) $1$. First,calculate the equivalent resistance of the parallel combinations:$R_{12} = \frac{R_1 	imes R_2}{R_1 + R_2} = \frac{2 	imes 2}{2 + 2}

Vedclass · Accepted Answer

$I_4 = \frac{2}{5}\,A$ and $I_5 = \frac{8}{5}\,A$

Vedclass · Answer

(D) $1$. First,calculate the equivalent resistance of the parallel combinations:$R_{12} = \frac{R_1 	imes R_2}{R_1 + R_2} = \frac{2 	imes 2}{2 + 2} = 1\,\Omega$$R_{45} = \frac{R_4 	imes R_5}{R_4 + R_5} = \frac{20 	imes 5}{20 + 5} = \frac{100}{25} = 4\,\Omega$$2$. The total resistance of the circuit is the sum of the series components:$R_{eq} = R_{12} + R_3 + R_{45} + R_6 + r = 1 + 2 + 4 + 2 + 3 = 12\,\Omega$$3$. The total current $I$ flowing through the circuit is:$I = \frac{V}{R_{eq}} = \frac{24}{12} = 2\,A$$4$. Using the current divider rule for the parallel branch containing $R_4$ and $R_5$:$I_4 = I 	imes \frac{R_5}{R_4 + R_5} = 2 	imes \frac{5}{20 + 5} = 2 	imes \frac{5}{25} = 2 	imes \frac{1}{5} = \frac{2}{5}\,A$$I_5 = I - I_4 = 2 - \frac{2}{5} = \frac{10 - 2}{5} = \frac{8}{5}\,A$

As shown in the figure,a network of resistors is connected to a battery of $24\,V$ with an internal resistance of $3\,\Omega$. The currents through the resistors $R_4$ and $R_5$ are $I_4$ and $I_5$ respectively. The values of $I_4$ and $I_5$ are:

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