Switch S has been closed for a long time, and the electric circuit shown in Figu
ID: 1538218 • Letter: S
Question
Switch S has been closed for a long time, and the electric circuit shown in Figure P28.68 carries a constant current. Take C1 = 3.00 µF, C2 = 6.00 µF, R1 = 4.00 k, and R2 = 7.00 k. The power delivered to R2 is 2.80 W.
Switch s has been closed for a long time, and the electric circuit shown in Figure P28.68 carries a constant current. Take C1 4.00 k2, and R 7.00 k2. The power delivered to R is 2.80 W. 2 Figure P28.68 (a) Find the charge on C1 HCE (b) Now the switch is opened After many mi seconds, by how much has the charge on C2 changed? HC 3.00 HF C2 6.00 HF R1Explanation / Answer
With the switch closed, current exists in the circuit as shown in the top part of Figure. The capacitors carry no current.
For R2 we have
P = I2/R2 , then I = sqrt[ P/R2 ]= sqrt[ 2.40/7000] = 18.5 mA
The potential difference across R1 and C1 is
V1 = IR1 = 1.85x 10^2x 4000 = 74.1 V
The charge on C1 is
Q = C1V = 3.00 × 106 × 74.1 = 222 µC
The potential difference across R2 and C2 is:
V2 = IR2 = 1.85 × 102 × 7000 = 130 V
The charge on C2 is:
Q = C2V = 6.00 × 106 × 130 = 778 µC
The battery emf is:
V = V1 + V2 = 74.1 + 130 = 204 V
or V = IReq = I(R1 + R2) = 1.85 × 102 × (4000 + 7000) = 204 V
b)
In equilibrium after the switch has been opened, no current exists. The potential difference across each resistor is zero. The full 204 V appears across both capacitors. The new charge C2 is:
Q = C2V = 6.00 × 106 × 204 = 1222 µF
and the change is 1222 778 = 444 µF
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