Question

In the circuit given, the current through the zener diode is.

• A. 10 mA
• B. $6.67$ Ma
• C. 5 mA
• D. $3.33$ mA

Answer 1

Answer: (D)

$3.33$ mA

Answer 2

:

The voltage drop across $R_{2}$is

$V_{R_{2}} = V_{Z} =$10V

The current through $R_{2}$is

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

The voltage drop across $R_{1}$is ()

}{= 15V - 10V = 5V}$$ The current through $R_{1}$is $$I_{R_{1}} = \frac{V_{R_{1}}}{R_{1}} = \frac{5V}{500\Omega} = 10^{- 2}A = 10 \times 10^{- 3}A = 10mA$$ The current through the zener diode is $${I_{Z} = I_{R_{1}} - I_{R_{2}} }{= (10 - 6.67)mA = 3.33mA}$$