1.

The switch S has been closed for long time and the electric circuit shown caries a steady current. Let C_1 = 3.0muF, C_2 = 6.0 muF, R_1 = 4.0 kOmega, and R_1 = 7.0 k Omega The power dissipated in R_2 is 2.8 W.

Answer»

The power dissipated to the resistor `R_1 is 1.6W.`
The charge on capacitor `C_1 is 240muC`.
The charge on capacitor `C_2 is 440muC`.
Long time after switch is opened, the charge on `C_1 is 660 muC.`

SOLUTION :a., B., d.
`i = 2 xx 10^(-2) A `
`P_(R1) = i^2R_1 = (2xx10^(-2))^2 xx 4 xx 10^(3) = 1.6W`
`Q_(C1) = V_(R1) xx C_1 = 80 xx 3 xx 10^(-6) = 240 muC`
`Q_(C2)= V_(R2) xx C_2 = 140 xx 6 xx 10^(-6) = 840 muC`
`Q_(C1) = E_(C1) = 220 xx 3 xx 10^(-6) = 660 muC`


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