An 8 mu F capacitor is connected to a 6V battery and allowed to completely charg

Question

An 8 mu F capacitor is connected to a 6V battery and allowed to completely charge^1. a) How much charge does it hold? b) This capacitor (fully charged) is disconnected from the battery and is connected across a second initially uncharged capacitor. After the charge transfer is complete a voltmeter connected across the capacitor combination reads 2.2 Volts. Determine the capacitance of the second capacitor and the charge held by each capacitor in the combination. c) Now the pair of connected capacitors is connected across a third initially uncharged capacitor. After the charge transfer is complete, a voltmeter placed across the combination reads 1.43 Volts. Determine the capacitance of the third capacitor and the charge on each capacitor in the combination. d) Describe (in words) what will happen if the last combination is now connected across the battery while the combination is charged as in c. How much charge will each capacitor hold when fully charged by the battery, and how much charge (total) will pass thru the circuit while the capacitor combination charges? As discussed in class, the resistivity of a substance that conducts using a single type of charge carrier (e.g. just electrons or just one type of ion) is given by rho = m/q^2 n tau where m and q are properties of the individual charge carriers. Two vector quantities are also important in what follows, the drift velocity v_d = q/m E tau and the current density j = qnv_4.

Explanation / Answer

a) Q=C1×V =8×10^-6×6=48×10^-6C

b) let the charge held by C1 is Q1 and by C2 is Q2

Both the capacitor wil have same potential because charge wil flow until potential become same so both the capacitor are in prallael Q1+Q2=2.2×(C1+C2)........1

Q1+Q2=48×10^ -6

Using eq 1 we get C2= 48/2.2×10^-6-8×10^-6 =C2

C2=13.82×10^-6 F

Q1=C1V,Q2=C2V so Q1/Q2=C1/C2 =8/13.82=0.58

Q1+Q2=48×10^-6

0.58Q2+Q2=48×10^-6

1.58Q2=48×10^-6

Q2=48×10^-6/1.58=30.38×10^-6C so Q1 wil be 48×10-6-30.38×10^-6=17.62×10^-6C

C)V=1.43V similarly all are in parallel so Cnet =C1+C2+C3=8×10^-6+13.82×10^-6+C3........(2)

total charge wil remain same so 48×10^-6/1.43=Cnet

Cnet=33.6×10^-6

using eq 2 we get C3==11.78×110^-6F

Q1=C1V,Q3=C3V

Q1/Q3=C1/C3

Q3/Q1=C3/C1 =11.78×10^-6/8×10^-6=1.47

Q3=1.47Q1

similarly for Q2/Q1=13.82/8=1.73

Q1+Q2+Q3=48×10^-6

Q1+Q2+1.47Q1=48×10^-6

Q1+1.73Q1+1.47Q1=48×10^-6

4.2Q1=48×10^-6

Q1=11.43×0^-6 C

Q2=1.73 Q1 =1.73×11.43×10-6 =19.8×10^-6C

Q3=1.47×Q1=1.47×11.43×10^-6 =16.8×10^-6C

When the combination is connected across battery charge wil start flowing as all capacitors are not fully charged charge hold by each one isC1V ,C2V,C3V where V is potential of battery

Q=Q1+Q2+Q3 charge wil pass through circuit

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