You are using the circuit shown below to measure the capacitance C of an unknown

Question

You are using the circuit shown below to measure the capacitance C of an unknown capacitor. The voltage provided by the ideal DC power supply is V0 = 6.23 V, and the resistor in series with the capacitor has resistance R = 1.10 M. The ammeter (A) measures the current through the single loop circuit. You close the switch at t = 0, and the current in the loop has dropped to 40% of its initial value at t = 5.00 s.

(a) Suppose you assume that your ammeter is ideal. (An ideal ammeter has zero resistance and does not draw power from the circuit that it is testing.) Working under this assumption, what have you measured the capacitance C to be?
F

(b) After making your 'ideal ammeter' assumption in part (a), you recall that real ammeters have an internal resistance that ought to be taken into account. You test the ammeter and see that it has an internal resistance of RI = 198 k (not a very good ammeter!). What have you really measured the capacitance of the capacitor to be?
F

(c) How bad was the 'ideal ammeter' assumption that you made in part (a)? Answer this by calculating a percent error for your answer to part (a), using your answer to part (b) as the accepted value.
%

(d) "How important is it that you consider the internal resistance of the ammeter?" The answer to this question depends on the error that you are willing to tolerate. For example, what is the largest value of RI for which the 'ideal ammeter' measurement of C is no more than 10% away from the 'real ammeter' measurement of C?
k

Explanation / Answer

a) I = I0 [ e^(-t /RC )]

0.40I0 = I0 e^(-t/RC)

-t/RC = ln(0.40) = 0.916

5 = 0.916 x 1.1 x 10^6 x C

C = 4.96 x 10^-6 F   = 4.96 uF

b) Rnet = 1.1 x 10^6 + 198 x10^3 = 1.298 x 10^6 ohm

0.40I0 = I0 e^(-t/RC)

-t/RC = ln(0.40) = 0.916

5 = 0.916 x 1.298 x 10^6 x C

C = 4.205 x 10^-6 F   = 4.205 uF

C) % error = (| 4.96 - 4.205 | / 4.205) x 100

            = 17.95 %

D) suppose real capacitance is C then

(4.96 - C) / C = 0.10

C = 4.509 uF

-t/RC = ln(0.40) = 0.916

5 = 0.916 x R x 10^-6 x 4.509

R = 1.2102 x 10^6 ohm

R = 1.1 x 10^6 + r = 1.2102 x 10^6

r = 1.1 x 10^6 ohm

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