330 Physics Laboratory Manual Layd Figure 33-1 Simple series RC circuit. in the

Question

330 Physics Laboratory Manual Layd Figure 33-1 Simple series RC circuit. in the discharging case will be in the opposite direction from the current in the charging case but the magnitude of the current is the same in both cases. Consider the circuit shown in Figure 33-2 consisting of a power supply of emf e, a capacitor C, a switch S, and a voltmeter with an input resistance of R. If initially the switch S is closed, the capacitor is charged almost immediately to E, the voltage of the power supply. When the switch is open discharges through the resistance of the meter R with a time constant given by RC. With the switch open, the only elements in the circuit are the capacitor C and the voltmeter resistance R, and thus the voltage across the capacitor is equal to the voltage across the voltmeter. It is given by (Eq. 3) Rearranging and taking the natural log of both sides of the equation gives In(e/V) = (1/RC)t Eq. 4) If the voltage across the capacitor is determined as a function of time, a graph of Infe/) versus t will give a straight line with a slope of (1/RC). Thus RC can be determined, and if R the voltmeter resistance is known, then C can be determined. If an unknown resistor is placed in parallel with the voltmeter, it produces a circuit like that shown in Figure 33-3. The capacitor can again be charged and then discharged, but now the time constant will be equal to R(C where R, is the parallel combination of R and Ru. If the relationship between R, Ru, and solved for Ru, the result is Ru Eq. 5) Therefore, a measurement of the capacitor voltage as a function of time will produce a dependence like that given by Equation 4, except that the slope of the straight line will be (1/R,C). Thus if C is known and R,C is found from the slope, then R can be determined. Using Equation 5, Ru can be found from R and R Power Supply C Capacitor Voltmeter Figure 33-2 An RC circuit using a voltmeter as the resistance.

Explanation / Answer

Given,

C = 10 x 10-6 F

V = 12 V

R = 10 x 106 Ohm

T = 35 sec

V= Ee-t/RC = 12 x e-35/(10 x 10^6 x 10 x 10^-6) = 12 x 0.705 = 8.45 V

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