FInd the following (Let C1=26.80 microF and C2=20.80 microF): a) the equivalent

Question

FInd the following (Let C1=26.80 microF and C2=20.80 microF):

a) the equivalent capacitance of the capacitors in the figure above ___________microF.

b) the charge on each capacitor:

on the right 26.80 microF capacitor________microC.

on the left 26.80 microF capacitor_______microC.

on the 20.80 microF capacitor________microC.

on the 6.00 microF capacitor_________microC.

c) the potential differences across each capacitor

on the right 26.80 microF capacitor_______________V

on the left 26.80 microF capacitor _________________V

on the 20.80 microF capacitor__________V

on the 6.00 microF capacitor__________V

Explanation / Answer

Let the capacitance of capacitors C1 = 26.8 µF

Let the capacitance of the capacitor C2 = 20.80 µF

Now let the capacitor with capacitance 6 µF be C3

Voltage supplied by the source is V = 9V

a)Firstly we can observe that C2 and C3 are in parallel, so we can write their equivalent capacitance as

           C23 = C2 + C3 = 20.80 µF + 6 µF = 26.80 µF

Now the capacitors C1 , C23 and C1 will be in series

The equivalent capacitance of the three capacitors is

    1 / C = 1 / C1 + 1 / C23 + 1 / C1

             = ( 1 / 26.8µF + 1/ 26.8µF + 1/ 26.8µF )

             = 3 / 26.8 µF

Then C = 26.8 µF / 3 = 8.933 µF

So the equivalent capacitance of the circuit is C = 8.933 µF

b)We have to start from backwards to calculate the charges and voltages at various capacitors

Total charge Q = C * V = 8.933 µF * 9 V = 80.39 µC

In series total charge on each capacitor is same

So charge on capacitor C1 on right side is Q1 = 80.39 µC = 80.4 µC

Then voltage is given by V1 ( right ) = Q1 / C1 = 80.4 µC / 26.8µF = 3 V

Similarly voltage on C23 = Q23 / C23 = 80.4 µC / 26.8µF = 3 V

Now we know that C2 and C3 are in parallel , the voltage along these capacitors is same i.e V23 = V2 = V3 = 3 V

So the charge on capacitor C2 is Q2 = C2 * V2 = 20.80 µF * 3 V = 62.4 µC

Similarly charge on C3 is Q3 = C3 * V3 = 6 µF * 3 V = 18 µC

From the above calculations we understand that voltage across C1 ( right ) , C23 is 3 V each

Then V = V1 ( left ) + V23 + V1 ( right )

          9 = V1 ( left ) + 3 + 3

Finally V1 = 3 V

So the charge on the capacitor C1 ( left ) is Q1 = C1 * V1 = 26.8 µF * 3 V = 80.4 µC

So the voltage across each capacitor is equal to 3V

Charge on capacitor 1 ( left ) is Q1 = 80.4µC

Charge on capacitor 2 is Q2 = 62.4 µC

Charge on capacitor 3 is Q3 = 18µC

Charge on capacitor 1 ( right ) is Q1 = 80.4 µC

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.