Question

A 12 volt supply is in series with a resistor and an LDR. In light, the LDR has a potential difference of 3.0 volts and in the dark it has a potential difference of 6.0 volts. Find the ratio of the series resistor to the resistance of the LDR in light.  

Answer 1

Assuming there is no other resistance in the circuit

Sum of p.d. across LDR and resistor is$12V$

$V\propto\>R$

Where $V=$ voltage

$R=Resistance$

In light p.d across LDR$=3V$

$\therefore\>p.d$ across R $=12-3$ $=9V$

Ratio of voltage $=9:3$

$=3:1$

$\implies$ Ratio of resistance $=3:1$