This question appeared in parts in TYS unfortunately. It is a interesting question.
di) Find effective resistance of 600 ohms and 3000 ohms which is 500 ohms. The main current is 12 V divide by the total resistance of (500 + 30) ohms
The current throught LDR is p.d /R where p.d = (0.0226 x 500). Hence the current through LDR is 11.3/3000 = 3.77x10-3 A.
The power dissipated in LDR = (3.77mA2 x 3000)= 0.043 W
When exposed to sun, the resistance falls to 100 ohms, repeat calculation.
The power is 0.67 W . The LDR will be damdage as it is above the marked 0.5 W.
To calculate the pd across LDR . connet the 2 points across LDR to the potentiometer to replace the second e.m.f (usually 'to be measure'.Draw the potentiometer first and below the circuit shown with LDR . The connection between the two circuits must include a galvanometer and the jockey.
