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(A) First,we check if the Zener diode is in the breakdown region. Assume the Zener diode is not conducting $(i_Z = 0)$.The circuit becomes a series co

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(A) First,we check if the Zener diode is in the breakdown region. Assume the Zener diode is not conducting $(i_Z = 0)$.The circuit becomes a series combination of $R_1 = 500 \ \Omega$ and $R_2 = 1500 \ \Omega$ in parallel with another $R_2 = 1500 \ \Omega$ (as shown in the diagram,the load resistor is $1500 \ \Omega$).The voltage across the parallel combination is $V_p = 12 	imes \frac{750}{500 + 750} = 12 	imes \frac{750}{1250} = 7.2 \ V$.Since $7.2 \ V Therefore,the current through the Zener diode is $0 \ mA$.

In the given circuit,the current through the Zener Diode is close to.......$mA$.

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