JEE MAIN - Physics (2021 - 16th March Morning Shift - No. 19)
The value of power dissipated across the zener diode (Vz = 15V) connected in the circuit as shown in the figure is x $$\times$$ 10$$-$$1 watt.
_16th_March_Morning_Shift_en_19_1.png)
The value of x, to the nearest integer, is __________.
The value of x, to the nearest integer, is __________.
Answer
5
Explanation
_16th_March_Morning_Shift_en_19_2.png)
$$i = {7 \over {35}} = {1 \over 5}A$$
$${i_1} = {{15} \over {90}} = {1 \over 6}A$$
$${i_2} = i - {i_1}$$
$${i_2} = {1 \over 5} - {1 \over 6}$$
$${i_2} = {1 \over {30}}A$$
Power across diode; P = V2 i2
$$P = 15 \times {1 \over {30}}$$
P = 0.5 W
$$ \therefore $$ P = 5 $$\times$$ 10$$-$$1 W
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