A battery of internal resistance 4,Omega is c | vedclass

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(A) According to the maximum power transfer theorem,the maximum power is delivered to the external circuit when the external resistance $(R_{eq})$ is

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$19/21\,\Omega$

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(A) According to the maximum power transfer theorem,the maximum power is delivered to the external circuit when the external resistance $(R_{eq})$ is equal to the internal resistance $(r)$ of the battery.$1$. First,simplify the circuit. The circuit consists of two parallel branches connected across the battery terminals $A$ and $B$.$2$. The left branch has resistors $2R$,$4R$,and $6R$ in series. Their equivalent resistance is $R_1 = 2R + 4R + 6R = 12R$.$3$. The right branch has resistors $R$,$2R$,and $4R$ in series. Their equivalent resistance is $R_2 = R + 2R + 4R = 7R$.$4$. These two branches are in parallel. The equivalent external resistance $R_{eq}$ is given by:$\frac{1}{R_{eq}} = \frac{1}{12R} + \frac{1}{7R} = \frac{7 + 12}{84R} = \frac{19}{84R}$$R_{eq} = \frac{84R}{19}$$5$. For maximum power transfer,$R_{eq} = r$,where $r = 4\,\Omega$.$\frac{84R}{19} = 4$$84R = 76$$R = \frac{76}{84} = \frac{19}{21}\,\Omega$.

$A$ battery of internal resistance $4\,\Omega$ is connected to a network of resistors as shown in the figure. If maximum power is to be delivered to the network,the magnitude of $R$ in $\Omega$ should be:

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