Modern Physics Ques 219
- For the circuit shown below, the current through the Zener diode is
(Main 2019, 10 Jan II)
(a) $14$ $ mA$
(b) zero
(c) $5 $ $mA$
(d) $9 $ $mA$
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Answer:
Correct Answer: 219.(d)
Solution:
- In the circuit, let the current in branches is as shown in figure below
By Kirchhoff’s node law,
$ I _1=I _2+I _3 \quad …….(i) $
Now, when diode conducts, voltage difference between points $A$ and $B$ will be
$ V _{A B}=120-50=70$ $ V $
So, $\quad$ current $I _1=\frac{V _{A B}}{5 k \Omega}=\frac{70}{5 \times 10^{3}}$
$ I _1=14 $ $mA \quad …….(ii) $
Since, diode and $10 $ $k \Omega$ resistor are in parallel combination, so voltage across $10 $ $k \Omega$ resistor will be $50 $ $V$ only.
$ \begin{array}{ll} \Rightarrow & I _3=\frac{50}{10 k \Omega}=\frac{50}{10 \times 10^{3}} \\ \Rightarrow & I _3=5 mA \quad …….(iii) \end{array} $
$\therefore$ From Eqs. (i), (ii) and (iii), we get
$ 14$ $ mA=I _2+5 $ $mA $
or current through diode,
$ I _2=14 $ $mA-5 $ $mA=9 $ $mA $
BETA
