For the circuit shown in 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, magnitude of the power dissipated in `R_(L)` will decreases by a factor `9`

Answer

A, D

Explanation

`R_(eq)=2 k Omega+(6k Omega||1.5 k Omega)`
`R_(eq)=3.2 k Omega`
(A) `I=(24V)/(3.2 k Omega)impliesI=7.5 mA`
`(B) V_(R_(L))=1.2/3.2xx24impliesV_(R_(L))=9 V`
`V_(1)=24-9impliesV_(2)=15` volt

(C) `P_(R_(1))=((15)^(2))/((2xx10^(3))=225/(2xx10^(3))` Watt
`P_(R_(L))=((9)^(2))/(6xx10^(3))=81/6` watt
(D) In first case
`P_(R_(L))=((9)^(2))/(1.5xx10^(3))`
For second case
`V'_(R_(L))=((6//7))/((6+6//7))xx24 V=3 V`
`P'_(R_(L))=((3)^(2))/((1.5xx10^(3)))` Watt

`P'_(R_(L))=1/9 P_(R_(L))`