Question

A Zener diode is connected to a battery and a load as shown below: The current $I,I_Z$ and $I_L$ are respectively :

• A. $12.5mA,735mA,5mA$
• B. $12.5mA,5mA,7.5mA$
• C. $15mA,7.5mA,7.5mA$
• D. $15mA,5mA,10mA$

Answer 1

The correct option is B $12.5mA,5mA,7.5mA$

Given that,

$R_L=2k\Omega$

$R=4k\Omega$

$V=60V$

$V_Z=10V$

The Zener diode in the breakdown region maintain a constant voltage, the voltage across $R_L$ is

$V_{R_L}=V_L=10V$

Now, the load current is

$I_L=\frac{V_L}{R_L}$

$I_L=\frac{10}{2000}$

$I_L=5mA$

Now, using Kirchhoff’s law

$I_L{R}_L-60+4000\Omega I=0$

$10-60+4000I=0$

$4000I=50$

$I=\frac{50}{4000}$

$I=0.0125$

$I=12.5mA$

Now, from Kirchhoff’s current law

$I=I_Z+I_L$

$I_Z=12.5-5$

$I_Z=7.5mA$

Hence, the current is $12.5mA$, $5mA$ and $7.5mA$