JEE MAIN - Physics (2018 - 16th April Morning Slot - No. 25)
In the given circuit, the current through zener diode is :
_16th_April_Morning_Slot_en_25_1.png)
5.5 mA
6.7 mA
2.5 mA
3.3 mA
Explanation
Voltage across R1 is given by $${V_{{R_1}}} = V - {V_Z} = 15 - 10 = 5\,V$$.
Resistance $${R_1} = 500\,\Omega $$
_16th_April_Morning_Slot_en_25_2.png)
Therefore, current through R1 is given by
$${I_{{R_1}}} = {{5V} \over {500\,\Omega }} = 0.01\,A$$
Now, current through R2 is given by
$${I_{{R_2}}} = {{10V} \over {1500\,\Omega }} = 0.067\,A$$
Therefore, the current through Zener diode will be
$$0.1 - 0.0067 = 0.0033\,A = 3.3mA$$
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