For the circuit shown in the figure

Options

A. the current I through the battery is 7.5 mA
B. the potential difference across `R_L` is 18 V
C. ratio of powers dissipated in `R_1 and R_2` is 3
D. If `R_1 and R_2` are interchanged, the magnitude of the power dissipated in `R_L` will decrease by a factor of 9

Explanation

`R_(total)=2+(6 times 1.5)/(6+1.5)=32 k Omega`
(a) `I=(24V)/(32 k Omega)=7.5 mA =I_(R1) implies I_(R_2)=(R_L/(R_L+R_2))I implies IR_2=1.5/7.5 times 7.5=1.5mA`
(b) `V_(R_1)= (J_(R_L) (R_L)=9V` ( c) `P_(R_1)/P_(R_2)=((r_(R1)^2)R_1)/((r_(R2)^2)R_2)=((7.5)^2 (2))/((1.5)^2 (6) =25/3`
When `R_1 and R_2` are inter changed, then `(R_2R_L)/(R_2+R_L)=(2 times 1.5)/3.5=6/7k Omega`
Now potential difference across `R_L` will be `V_L=24 [(6//7)/(6+6//7)]=3V`.Earlier it was 9V
SInce `P=V^2/R or P prop V^2`
In new situation potential difference has been decreased three times. Therefore, power dissipated will decreases be a factor 9.