Use the exact values you enter to make later calculations. Two capacitors are co

Question

Use the exact values you enter to make later calculations.

Two capacitors are connected in series. Using the same instruments as you did in the lab, you record both the capacitance of and the voltage difference across the electrodes of the capacitors individually. Complete the table below using theoretical formulae.

Use the exact values you enter to make later calculations.

Three capacitors C1 = 4.5 µF, C2 = 9.5 µF, and C3 = 21.5 µF are connected as shown with a battery of voltage V = 18 V. (When entering units, use micro for the metric system prefix µ.)

(a) What is the equivalent capacitance of this circuit?

(b) What is the charge on C1?

(c) What is the voltage across each capacitor?

(d) What is the charge on C2 and C3?

Capacitor #1 (Larger Capacitance) Capacitor #2 (Smaller Capacitance) Total Arrangement (Theoretical) Voltage Difference (V) 3.430 6.500 ? Capacitance (µF) 1004.000 545.000 ? Charge (µC) ? ? ? Energy (µJ) ? ? ?

Explanation / Answer

a)
Here C2 and C3 are connected in parallel.

C23 = C2 + C3

= 9.5 + 21.5

= 31 micro F

This sombination is in series with C1.

so, equivalent capacitance,

Cnet = C1*C23/(C1+C23)

= 4.5*31/(4.5+31)

= 3.93 micro F or 3.93*10^-6 F

b) V1 = V*C23/(C1+C23)

= 18*31/(4.5+31)

= 15.72 volts

V2 = V3 = V*C1/(C1+C23)

= 18*4.5/(4.5+31)

= 2.28 volts

c) Q2 = C2*V2

= 9.5*2.28

= 21.66 micro C or 21.66*10^-6 c

Q3 = C3*V3

= 21.5*2.28

= 49.02 micro C or 49.02*10^-6 C

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