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(C) The circuit consists of two branches connected in parallel. The upper branch has resistors $20 \, \Omega$ and $4 \, \Omega$ in series, so its resi

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$1$

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(C) The circuit consists of two branches connected in parallel. The upper branch has resistors $20 \, \Omega$ and $4 \, \Omega$ in series, so its resistance is $R_1 = 20 + 4 = 24 \, \Omega$.The lower branch has resistors $50 \, \Omega$ and $10 \, \Omega$ in series, so its resistance is $R_2 = 50 + 10 = 60 \, \Omega$.The total current $I = 1.4 \, A$ divides into these two parallel branches.Using the current divider rule, the current $I_1$ through the upper branch (which contains the $4 \, \Omega$ resistor) is given by:$I_1 = I 	imes \frac{R_2}{R_1 + R_2}$$I_1 = 1.4 	imes \frac{60}{24 + 60}$$I_1 = 1.4 	imes \frac{60}{84}$$I_1 = 1.4 	imes \frac{5}{7} = 0.2 	imes 5 = 1 \, A$.Thus, the current through the $4 \, \Omega$ resistor is $1 \, A$.

In the given figure, the current through the $4 \, \Omega$ resistor is ................. $A$.

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