Modern Physics 6 Question 12
13. 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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Solution:
- In the circuit, let the current in branches is as shown in figure below
By Kirchhoff’s node law,
$$ I _1=I _2+I _3 $$
Now, when diode conducts, voltage difference between points $A$ and $B$ will be
$$ V _{A B}=120-50=70 V $$
So, $\quad$ current $I _1=\frac{V _{A B}}{5 k \Omega}=\frac{70}{5 \times 10^{3}}$
$$ I _1=14 mA $$
Since, diode and $10 k \Omega$ resistor are in parallel combination, so voltage across $10 k \Omega$ resistor will be $50 V$ only.
$$ \begin{array}{ll} \Rightarrow & I _3=\frac{50}{10 k \Omega}=\frac{50}{10 \times 10^{3}} \\ \Rightarrow & I _3=5 mA \end{array} $$
$\therefore$ From Eqs. (i), (ii) and (iii), we get
$$ 14 mA=I _2+5 mA $$
or current through diode,
$$ I _2=14 mA-5 mA=9 mA $$
