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(C) The $10 \Omega$ and $40 \Omega$ resistors are connected in parallel. Let $i_1$ be the current through the $10 \Omega$ resistor and $i_2$ be the cu

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$8$

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(C) The $10 \Omega$ and $40 \Omega$ resistors are connected in parallel. Let $i_1$ be the current through the $10 \Omega$ resistor and $i_2$ be the current through the $40 \Omega$ resistor.Since they are in parallel,the potential difference across them is the same:$i_1 	imes 10 = i_2 	imes 40$Given $i_1 = 2.5 	ext{ A}$,we have:$2.5 	imes 10 = i_2 	imes 40$$i_2 = \frac{25}{40} = 0.625 	ext{ A}$At the junction point,the total current $i$ flowing through the circuit is:$i = i_1 + i_2 = 2.5 + 0.625 = 3.125 	ext{ A}$The equivalent resistance $R'$ of the parallel combination of $10 \Omega$ and $40 \Omega$ is:$R' = \frac{10 	imes 40}{10 + 40} = \frac{400}{50} = 8 \Omega$The total resistance of the circuit is $R'' = (R + 8) \Omega$.Using Ohm's law,$V = i 	imes R''$:$50 = 3.125 	imes (R + 8)$$R + 8 = \frac{50}{3.125} = 16$$R = 16 - 8 = 8 \Omega$Thus,the value of $R$ is $8 \Omega$.

In the given circuit,if the current through the $10 \Omega$ resistor is $2.5 \text{ A}$,the value of $R$ is . . . . . . . (in $\Omega$)

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