For the circuit shown in figure

Options

A. the current I through the batter is 7.5 mA.
B. the potential difference across R_L is 18V.
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 decrease by a factor of 9.

Answer

A, D

Explanation

(a,d)
`R_p = ((R_2 xx R_L)/(R_2 + R_L)) = (6 xx 1.5)/(6+1.5) = 9/(7.5)kOmega`
`:. I = (24/(9/7.5 + 2)) mA = (24 xx 7.5)/24 = 7.5 mA.`
`rArr option (a) is correct.
The potential difference across `R_L = potential`
difference across `R_p`
`=(7.5mA)(9/7.5 kOmega) = 9V`.
`rArr option (b) is incorrect.
Now, (Power dissipation across `R_1`)/(Power dissipation across `R_2`) = `((15)^2/2/(9)^2/6)`
`= (15 xx 15)/ 2 xx (6/(9 xx 9)) = 8.33`
`rArr option (c) is incorrect.
The magnitude of power dissipated accross `R_2` is
`(9)^2. (1.5).
Now when `R_1 and R_2` are interchanged the equivalent
resistance between `R_1 and R_L = (2xx1.5)/(2+1.5) = 3/3.5 kOmega`

`:. Potential drop across this equivalent resistance.
`((3/3.5)/(3/3.5 +6)) xx 24 = 3/24 xx 24 = 3V.`
`:. Potential difference accross `R_L = 3^2/(1.5) = 1/9 [9^2/(1.5)]`
`:. The magnitude of the power dissipation in `R_L` will
decrease by a factor 9 if `R_1 and R_2` are intercharged.
(d) is the correct option.