A zener diode having $V_z$ = 12 V and $I_{Z(max)}$ = 40 mA is used in a voltage regular circuit with unregulated power supply giving E = 19 V as shown in the circuit diagram shown below. The minimum value of the series resistor R required when the load resistance of $400 \Omega$ is connected across the zener diode, so that the zener should not burn out, will be:

$100 \Omega$

The correct answer is Option (1) → $100 \Omega$

According to information given, the potential at various point of circuit as represented below.

Here $l_1=\frac{12}{400}=\frac{3}{100}$ = 30 mA

$l_2$ = 40 mA    (Given)

∴   $l$ = 70 mA

Now, $R = \frac{19-12}{70 mA}=\frac{7}{70 \times 10^{-3}}$

$R=100 \Omega$