The current I in the network shown in the fig | vedclass

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(B) From the circuit diagram, the three $6 \, \Omega$ resistors are connected in parallel to the voltage source, and one $6 \, \Omega$ resistor is in

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$3$

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(B) From the circuit diagram, the three $6 \, \Omega$ resistors are connected in parallel to the voltage source, and one $6 \, \Omega$ resistor is in series with the battery.Let the three parallel resistors be $R_1, R_2, R_3$ where $R_1 = R_2 = R_3 = 6 \, \Omega$.The equivalent resistance of these three parallel resistors is $R_p = \frac{6}{3} = 2 \, \Omega$.This equivalent resistance $R_p$ is in series with the remaining $6 \, \Omega$ resistor.Therefore, the total equivalent resistance of the circuit is $R_{net} = R_p + 6 \, \Omega = 2 \, \Omega + 6 \, \Omega = 8 \, \Omega$.The voltage of the battery is $V = 24 \, V$.Using Ohm's law, $I = \frac{V}{R_{net}} = \frac{24 \, V}{8 \, \Omega} = 3 \, A$.Thus, the current $I$ is $3 \, A$.

The current $I$ in the network shown in the figure is .......... $A$.

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