JEE MAIN - Physics (2019 - 8th April Morning Slot - No. 21)
The reverse breakdown voltage of a Zener
diode is 5.6 V in the given circuit.
The current IZ through the Zener is :_8th_April_Morning_Slot_en_21_1.png)
The current IZ through the Zener is :
7 mA
17 mA
15mA
10 mA
Explanation
9 = Vz + VR1
VZ = 5.6 V
VR1 = 9 - 5.6
VR1 = 3.4
$${I_{{R_1}}} = {{{V_{{R_1}}}} \over R} = {{3.4} \over {200}};\,{I_{{R_1}}} = 17\,mA$$
Vz = VR2 = IR2 (R2 )
$${{5.6} \over {700}} = {I_{{R_2}}}$$; $${I_{{R_2}}} = 7\,mA$$
Iz = (17 – 7) mA = 10 mA
VZ = 5.6 V
VR1 = 9 - 5.6
VR1 = 3.4
$${I_{{R_1}}} = {{{V_{{R_1}}}} \over R} = {{3.4} \over {200}};\,{I_{{R_1}}} = 17\,mA$$
Vz = VR2 = IR2 (R2 )
$${{5.6} \over {700}} = {I_{{R_2}}}$$; $${I_{{R_2}}} = 7\,mA$$
Iz = (17 – 7) mA = 10 mA
Comments (0)
