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(A) First,calculate the equivalent resistance of the three $3\, k\Omega$ resistors connected in parallel:$\frac{1}{R_p} = \frac{1}{3} + \frac{1}{3} +

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(A) First,calculate the equivalent resistance of the three $3\, k\Omega$ resistors connected in parallel:$\frac{1}{R_p} = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = \frac{3}{3} = 1 \implies R_p = 1\, k\Omega$.Now,the circuit consists of the $5\, k\Omega$ resistor,the equivalent resistance $R_p = 1\, k\Omega$,and the internal resistance of the battery $r = 1\, k\Omega$ all in series.The total resistance $R_{eq} = 5\, k\Omega + 1\, k\Omega + 1\, k\Omega = 7\, k\Omega$.The total current $I$ flowing through the circuit is given by Ohm's law:$I = \frac{V}{R_{eq}} = \frac{21\, V}{7\, k\Omega} = 3\, mA$.Since the $5\, k\Omega$ resistor is in series with the rest of the circuit,the same current $I = 3\, mA$ flows through it.

In the figure given,the electric current flowing through the $5\, k\Omega$ resistor is $'x' \, mA.$
The value of $x$ to the nearest integer is .........

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In the figure given,the electric current flowing through the $5\, k\Omega$ resistor is $'x' \, mA.$The value of $x$ to the nearest integer is .........