For the circuit shown in figure

Options

A. the current `I` through the battery is `7.5 mA`
B. the potential diifference across `R_(3)` is 18 V
C. ratio of a power 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_(3)` will decrease by a factors of `9`

Answer

a,d

Explanation

Total resistance of circuit
`= R_(1) + (R_(2)R_(3))/(R_(2) + R_(3)) = 2 + (6xx1.5)/(6+1.5)`
`= 2+ 1.2 = 3.2 k Omega`
(a) Current through the battery,
`I = (24V)/(3.2 k Omega ) = (24)/(3.2 xx 1000) = 7.5mA`
(b) Pot diff across `R_(2) and R_(3)`
`V_(2) = (7.5mA) xx (1.2 k Omega) = 9V`
(c) Pot diff across `R_(1)`
`V_(1) = 7.5mA xx 2k Omega = 15 V`
Power disspated across `R_(1) , P_(1) = (V_(1)^(2))/(R_(1)) = (15^(2))/(2 xx 10^(3))`
Power disspated across `R_(2) , P_(2) = (V_(2)^(2))/(R_(2)) = (9^(2))/(6 xx 10^(3))`
`:. (P_(1))/(P_(2)) = (15^(2) xx 6 xx 10^(3))/(2 xx 10 ^(3)xx9^(2)) = 8.3`
Power disspated across `R_(3)`,
`P_(3) = (9^(2))/(1.5 xx 10^(3)) = 54 mW`
(d) If `R_(1) and R_(2)` are interchanged total resistance
of circuit is `= 6 + (2xx 1.5)/(2+1.5) = 6+ (6)/(7) = (48)/(7) k Omega`
Total current, `I = (24 xx 7)/(48 xx 1000) = 3.5 mA`
pot diff across `R_(1) = (3.5mA) xx (6)/(7) k Omega = 3V`
Power dissipated across `R_(3)`,
`P'_(3) = (3^(2))/(1.5 xx 10^(3)) = 6 mW`
As `P'_(3) lt P_(3)`, power dissipation in `R_(3)` will decrease
`:. (P_(3))/(P'_(3)) = (54)/(6) = 9` (decrease)