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(D) The circuit consists of two parallel blocks connected in series.First, calculate the equivalent resistance of the first block $(R_1)$ containing t

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$2\, A$ and $2\, A$

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(D) The circuit consists of two parallel blocks connected in series.First, calculate the equivalent resistance of the first block $(R_1)$ containing two $4\, \Omega$ resistors in parallel:$R_1 = \frac{4 	imes 4}{4 + 4} = 2\, \Omega$.Next, calculate the equivalent resistance of the second block $(R_2)$ containing two $6\, \Omega$ resistors in parallel:$R_2 = \frac{6 	imes 6}{6 + 6} = 3\, \Omega$.The total equivalent resistance of the circuit is $R_{eq} = R_1 + R_2 = 2\, \Omega + 3\, \Omega = 5\, \Omega$.The total current flowing through the circuit is $I = \frac{V}{R_{eq}} = \frac{20\, V}{5\, \Omega} = 4\, A$.Since the two $4\, \Omega$ resistors are in parallel, the current $I$ splits equally between them. Thus, the current through each $4\, \Omega$ resistor is $I_4 = \frac{4\, A}{2} = 2\, A$.Similarly, since the two $6\, \Omega$ resistors are in parallel, the current $I$ splits equally between them. Thus, the current through each $6\, \Omega$ resistor is $I_6 = \frac{4\, A}{2} = 2\, A$.Therefore, the current flowing through a $4\, \Omega$ resistance and a $6\, \Omega$ resistance is $2\, A$ and $2\, A$ respectively.

Four resistances are connected in a circuit as shown in the given figure. The electric current flowing through a $4\, \Omega$ resistance and a $6\, \Omega$ resistance is respectively:

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