The circuit in the figure below has been connected for a long time. Let R 1 = 8.

Question

The circuit in the figure below has been connected for a long time. Let R1 = 8.30 and R2 = 5.90 .

(a) What is the potential difference across the capacitor?

(b) If the battery is disconnected from the circuit, over what time interval does the capacitor discharge to one-fifth its initial voltage?

The original circuit of the figure above has been redrawn into parts of four equivalent circuits in the diagrams (i) through (iv).

(a) After a long time interval, the capacitor branch will carry negligible current. The current is shown in diagram (i). To find the voltage at point a, we first find the current. Using the voltage rule, we have

For the right-hand branch,

Thus, the voltage across the capacitor is

Va Vb =  V   V =  V.

Suppose the battery is removed leaving an open circuit. We are left with the circuit in diagram (ii), which can be reduced to equivalent circuits represented in diagrams (iii) and (iv). From diagram (iii), we see that the two resistances are in parallel so that the equivalent resistance in diagram (iv) is given by the following.

Riv =

=  

The capacitor discharges through this equivalent resistance

Riv.

According to the equation for a capacitor discharged through a resistor R,

q = Qet/RC,

we multiply both sides by the capacitance to give the following in terms of the initial voltage

Vi

and the voltage

V

after discharge.

V = Vie-t/RC

Note that the resistance in the equation above in this circuit is the equivalent resistance we calculated for

Riv

in diagram (iv). Solving for the capacitor's discharge time t, we have the following.

10.0 V (1.00 )I2 R2I2 = 0

I2 =  A

Va V0 = R2I2 =  V. 10.0 V 1.00 1.00 uF 2.00 (2

Explanation / Answer

1) a) Notice that when the capacitor is "fully" charged (only then), the 1 and 5.9 are connected in series. We can relate the potential at junction B with the potential at junction A.

VA - VD = I2*5.9 = 10V*5.9/(5.9 + 1) = 8.55 V

Similarly VC - VD = I3*2 = 10V*2/(2 + 8.3) = 1.94 V

Therefore  VA - VC = (VA - VD) - (VC - VD) = 8.55 - 1.94 = 6.61 V

b) When the battery is disconnected, the capacitor is discharged through the system of resistors. A current of the same value will be flowing through the 1 and the 8.3 resistors. They are connected in series. Similarly, we can see that the 5.9 and 2 resistors are connected in series. Both pairs are connected in parallel at points A and C. The equivalent resistance of the system is therefore:

R = (1/(1+8.3) + 1/(5.9 + 2))-1 = 4.27 ohm.............. (1)

The loop rule applied to the equivalent circuit presented in the figure relates the rate of change in the voltage across the capacitor with the value of that voltage at any instant.

VC - R[-CdVc/dt] = 0 .............. (2)

We obtained a differential equation for function VC(t). Choosing the instant when the battery is disconnected as the reference instant t0=0s, we can express explicitly the time dependence of the voltage across the capacitor.

dVc/Vc = (-1/RC)dt

Vc(t) = Vc(0)*e^(-t/RC)

From this function we can write an equation for the time t5 at which the voltage discharges to a given value.

(1/5)Vc(0) = Vc(0)*e^(-t5/RC)

The solution to this equation is: t5 = -RC*ln(1/5) = -4.27*1*10^-6*(-1.6094) = 6.87*10^-6 s

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