12. A dielectric-filled capacitor is directly connected to a 12.0-V battery unti

Question

12. A dielectric-filled capacitor is directly connected to a 12.0-V battery until it is fully charged. Then the capacitor is disconnected from the battery and, once disconnected, the dielectric is removed so that air fills the capacitor instead. The dielectric used is pyrex glass, whose dielectric constant is 5.6, while the dielectric constant of air is 1.0059. (a) Determine the voltage across the capacitor plates after the dielectric is removed (b) With the dielectric removed, the capacitor is discharged by connecting it to a device with a resistance of 150 . Suppose that you would like the capacitor to discharge to 5.00% of its original voltage (the original voltage is the answer to part a) in 3.5 seconds. To achieve this, what should the capacitance be? Page 2 13. In the circuit shown below, ½ = 12.0 V, Ri = 2.00 , R2 = 10.0 , R3 = 5.00 , and R, = 8.00 . (a) Determine the current flowing through each resistor. Include the direction in which each current flows. (b) What is the total power consumed by the circuit? (c) If the resistors represent light bulbs, rank them in order of decreasing brightness. Rt Ra Rs 14. In the circuit shown below, all resistors have a resistance of 5.00 , while the battery's voltage is = 9.00 V. Determine (a) the current flowing through the battery, and (b) the power consumed by the cireuit

Explanation / Answer

ONly one question can be posted as per chegg guidelines

a)

With Battery disconnected charge remains constant

Qd=CdV =12Cd

Capaciance Without Dielectric

C=Cd(KPyrex/Kair)

Voltage across capacitor without dielectric

V =Qd/C =12Cd/Cd(Kpyrex/Kair)=12Kair/Kpyrex

V=12*1.0059/5.6 =2.14 Volts

b)

In a RC circuit charge as a function of time for discharging is given by

V=Voe-t/T

=>0.05Vo=Voe-3.5/T

Ln(0.05)=-3.5/T

T=1.168 s

Since T=RC

=>C=T/C =1.168/150

C=7.79*10-3F or 7.79 mF

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