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

Q.  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.
- Published on 20 Oct 15

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

ANSWER: Rs = 3 KΩ & Id = 0.15mA .0.19 mA
 
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,
Id = V / (Rs + RD )
= 40 / (3 + 200) x 103
= 0.19 x 10 -3 A
= 0.19 m A

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