Question

In the circuit diagram, the current through the Zener diode is :

• A. $10mA$
• B. $3.33mA$
• C. $6.67mA$
• D. $0mA$

Answer 1

Answer: (B)

The voltage drop across $R_2$ is

$V_{R_2}=V_z=10V$

The current through $R_2$ is

$I_{R_2}=\frac{V_{R_2}}{R_2}=\frac{10}{1500}=0.667\times {10}^{-2}A$

$=6.67\times {10}^{-3}A=6.67mA$

The voltage drop across $R_1$ is

$V_{R_1}=15V-V_{R_2}=15V-10V=5V$

The current through $R_1$ is

$I_{R_1}=\frac{V_{R_1}}{R_1}=\frac{5V}{500\Omega }={10}^{-3}A=10\times {10}^{-3}=10mA$

The current through Zener diode is

$I_z=I_{R_1}-I_{R_2}=(10-6.67)mA=3.33mA$