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(A) In the given circuit, the Zener diode is connected in parallel with the $4 	ext{ k}\Omega$ resistor. Since the Zener diode is operating in its br

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$\frac{2}{3} 	ext{ mA}$

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(A) In the given circuit, the Zener diode is connected in parallel with the $4 	ext{ k}\Omega$ resistor. Since the Zener diode is operating in its breakdown region, the voltage across it is constant at $6 	ext{ V}$.Because the Zener diode and the $4 	ext{ k}\Omega$ resistor are in parallel, the voltage across the $4 	ext{ k}\Omega$ resistor is also $6 	ext{ V}$.The total voltage in the circuit is $10 	ext{ V}$. The voltage drop across the $6 	ext{ k}\Omega$ series resistor is the difference between the source voltage and the Zener breakdown voltage:$V_{6	ext{k}\Omega} = 10 	ext{ V} - 6 	ext{ V} = 4 	ext{ V}$.The current flowing through the $6 	ext{ k}\Omega$ resistor is given by Ohm's law:$I = \frac{V_{6	ext{k}\Omega}}{R} = \frac{4 	ext{ V}}{6 	ext{ k}\Omega} = \frac{4}{6} 	ext{ mA} = \frac{2}{3} 	ext{ mA}$.

What will be the current flowing through the $6 \text{ k}\Omega$ resistor in the circuit shown, where the breakdown voltage of the Zener diode is $6 \text{ V}$?

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