Two capacitors with capacitances 3 mu F and 2 mu F are initially discharged. The

Question

Two capacitors with capacitances 3 mu F and 2 mu F are initially discharged. They are connected in series, and then the two ends of the combination are connected to a 10-V battery, as shown in Figure 1.24. The negative end of the battery is at zero potential. (a) How much charge docs the battery deliver? You can get started by figuring out the relationship between the charges in each capacitor and the voltage across each of them. (b) If we connect a single capacitor to the same battery, what would be the capacitance of the capacitor so that it draws the same charge from the battery? (c) Find the charges on each capacitor. (d) Find the voltage across each capacitor. (e) If point B is at zero potential, what is the potential at point A between the two capacitors?

Explanation / Answer

1

a)

C1 = 3 uF   , C2 = 2 uF

equivalent capacitance of the series combination is given as

Ceq = C1 C2/(C1 + C2) = 3 x 2 /(3 + 2) = 1.2 uF

V = potential difference = 10 volts

charge delivered is given as

Q = Ceq V = 1.2 x 10 = 12 uC

b)

C' = charge of single Capacitor

Q = 12 uC

V = voltage = 10 volts

Q = C' V

12 = C' (10)

C' = 1.2 uF

c)

since the capacitors are in series and in series same charge flows through each capacitor

hence charge on each capacitor = 12 uC

d)

Voltage across C1 is given as

V1 = Q/C1 = 12/(3) = 4 volts

Voltage across C2 is given as

V2 = Q/C2 = 12/(2) = 6 volts

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