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(B) $1$. Calculate the equivalent resistance of the parallel combination of $3\, \Omega$ and $6\, \Omega$ resistors:$\frac{1}{R_p} = \frac{1}{3} + \fr

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(B) $1$. Calculate the equivalent resistance of the parallel combination of $3\, \Omega$ and $6\, \Omega$ resistors:$\frac{1}{R_p} = \frac{1}{3} + \frac{1}{6} = \frac{2+1}{6} = \frac{3}{6} = \frac{1}{2} \implies R_p = 2\, \Omega$.$2$. Calculate the total resistance of the circuit:$R_{total} = R_p + 4.5\, \Omega + r_1 + r_2 = 2 + 4.5 + 0.5 + 1 = 8\, \Omega$.$3$. Calculate the net electromotive force $(EMF)$ of the circuit:The cells are connected in opposition, so $E_{net} = 8\, V - 4\, V = 4\, V$.$4$. Calculate the total current $(I)$ flowing in the circuit:$I = \frac{E_{net}}{R_{total}} = \frac{4\, V}{8\, \Omega} = 0.5\, A$.$5$. Use the current divider rule to find the current $(I_1)$ through the $3\, \Omega$ resistor:$I_1 = I 	imes \left( \frac{R_{other}}{R_1 + R_{other}} ight) = 0.5 	imes \left( \frac{6}{3+6} ight) = 0.5 	imes \frac{6}{9} = 0.5 	imes \frac{2}{3} = \frac{1}{3}\, A$.$6$. Given $I_1 = \frac{x}{3}\, A$, we have $\frac{x}{3} = \frac{1}{3}$, which implies $x = 1$.

In the circuit diagram shown in the figure below, the current flowing through the resistance $3\, \Omega$ is $\frac{x}{3}\, A$. The value of $x$ is $...........$

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