# 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 10^{3}

= 3 kΩ

Also, the dark current can be given by,

I_{d} = V / ( Rs + R_{D} )

= 40 / (3 + 200) x 10^{3}

= 0.19 x 10 ^{-3} A

= 0.19 m A