JEE MAIN - Physics (2021 - 20th July Morning Shift - No. 2)
For the circuit shown below, calculate the value of Iz :
_20th_July_Morning_Shift_en_2_1.png)
25 mA
0.1 A
0.15 A
0.05 A
Explanation
Consider the given figure and draw the direction of current as follows
_20th_July_Morning_Shift_en_2_2.png)
Here, current I is flowing through resistance Rs and I1 is the current flowing through R = 2000 $$\Omega$$
$$\therefore$$ Current, $$I = {{{V_i} - {V_z}} \over {{R_s}}} \Rightarrow I = {{100 - 50} \over {1000}}$$
= 50 mA .... (i)
[$$\because$$ Resistance and Zener diode are in parallel, therefore voltages across them are same]
and current, $${I_1} = {{50} \over {2000}} = 25$$ mA .... (ii)
By using Kirchhoff's current law,
$$\because$$ I = I1 + Iz
$$\Rightarrow$$ Iz = I $$-$$ I1 ..... (iii)
From Eq. (i) and (ii) and, we get
$$\Rightarrow$$ Iz = (50 $$-$$ 25) mA
Iz = 25 mA
Here, current I is flowing through resistance Rs and I1 is the current flowing through R = 2000 $$\Omega$$
$$\therefore$$ Current, $$I = {{{V_i} - {V_z}} \over {{R_s}}} \Rightarrow I = {{100 - 50} \over {1000}}$$
= 50 mA .... (i)
[$$\because$$ Resistance and Zener diode are in parallel, therefore voltages across them are same]
and current, $${I_1} = {{50} \over {2000}} = 25$$ mA .... (ii)
By using Kirchhoff's current law,
$$\because$$ I = I1 + Iz
$$\Rightarrow$$ Iz = I $$-$$ I1 ..... (iii)
From Eq. (i) and (ii) and, we get
$$\Rightarrow$$ Iz = (50 $$-$$ 25) mA
Iz = 25 mA
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