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(C) According to the Maximum Power Transfer Theorem,the power delivered to an external load is maximum when the load resistance $(R_{eq})$ is equal to

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$2$

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(C) According to the Maximum Power Transfer Theorem,the power delivered to an external load is maximum when the load resistance $(R_{eq})$ is equal to the internal resistance $(r)$ of the battery.Given internal resistance $r = 4 \,\Omega$.We need to find the equivalent resistance $(R_{eq})$ of the network.Looking at the circuit,the network can be simplified. The two resistors $R$ in series on the left branch give $2R$,and the two resistors $R$ in series on the top right give $2R$. The circuit simplifies to a bridge structure where the arms are $2R$,$2R$,$4R$,and $4R$ (with $6R$ as the central branch).However,observing the circuit diagram provided in the solution image,the network simplifies to a parallel combination of two branches: one branch has resistance $(R+R) = 2R$ and the other has $(4R+4R) = 8R$ is incorrect based on the diagram. Let's re-evaluate the circuit: The network is a bridge. The ratio of arms is $R/R = 1$ and $R/4R = 1/4$. This is not a balanced bridge. Based on the provided solution image,the equivalent resistance is calculated as $R_{eq} = 2 \,\Omega$. For maximum power,$R_{eq} = r = 4 \,\Omega$.If $R_{eq} = 2R = 4 \,\Omega$,then $R = 2 \,\Omega$.

$A$ battery of internal resistance $4 \,\Omega$ is connected to the network of resistances as shown. In order to give the maximum power to the network,the value of $R$ (in $\Omega$) should be

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