Numerical - Requisite series resistance and dark current for a relay which is under the control of photo-conductive cell

Compute the requisite series resistance and dark current for a relay which is under the control of photo-conductive cell with the illumination resistance of 2kΩ and dark resistance of 200 kΩ. Current supply to relay is about 8mA from the voltage supply of about 40 V under the illumination condition. It is also mandatory to de-energize the relay when the cell is in the midst of dark.

Options:

a. Rs = 2 KΩ & I_d = 0.15mA
b. Rs = 2 KΩ & I_d = 0.12 mA
c. Rs = 3 KΩ & I_d = 0. 10 mA
d. Rs = 3 KΩ & I_d = 0.15mA.0.19 mA

Correct Answer:d. Rs = 3 KΩ & I_d = 0.15mA.0.19 mA

Explanation:

Applying Ohm's law,

I = V / R
I = V / ( R+ r) since 'r' is a cell resistance
I = 40 / (R + r)
Therefore, we can write,

R = (40 / I) – r
When a photoconductive cell is illuminated,
R = ( 40/ 8x 10 ^-3 ) - 2x 10³
= 3 kΩ

Also, the dark current can be given by,
I_d = V / ( Rs + R_D )
= 40 / (3 + 200) x 10³
= 0.19 x 10 ^-3 A
= 0.19 m A