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For the capacitor network shown in Fig. E24.28, the potential difference across ab is 48 V. Find (a) the total charge stored in this network; (b) the charge on each capacitor; (c) the total energy stored in the network; (d) the energy stored in each capacitor; (e) the potential differences across each capacitor

Short Answer

Expert verified

(a) The total charge stored in the network is 3.2μC

(b) The charge on each capacitor is3.2μC

(c) Total energy stored in the network is3.14×10-5J and4.27×10-5J

(d) The potential difference across each capacitor is21.33V and26.67V

Step by step solution

01

Capacitance of a parallel plate capacitor:

We know that the capacitance of a parallel plate capacitor is given by:

C=QVab=ε0Ad

Where, C is the capacitance,Vab is the potential difference, Q is the magnitude of the charge on each plate, d is the distance between each plate, A is the area of the plate andε0 is the electric constant whose value is given by:8.85×10-12F/m

Step2: Calculating the equivalent capacitance

By solving the above equation, we get:

1Ceq=1C1+1C2

Now, calculating the equivalent capacitance

Ceq=C=C1C2C1+C2=150×10-9F120×10-9F150×10-9F+120×10-9F=66.6667×10-9F

Now, using the equation for C we get:

C=QVab

Putting the values and solving for we get:

Q=CVab=66.667×10-948=3.2μC

Since the capacitors are connected in series, we the equivalent capacitance is3.2μC for both

02

Calculating the potential energy stored

We that the potential energy stored in a capacitor U is equal to the work done while charging the capacitor:

U=W=VQ2=Q22C

Substituting the values, we get:

U=3.2×10-622×66.667×10-9=7.67×10-5J

Therefore, the energy stored is7.67×10-5J

03

Energy stored in each capacitor

U1=Q22C1=3.2×10-622×150×10-9=3.41×10-5J

Similarly for the second capacitor C2, stored energy is: 4.27×10-5J

For potential difference we will use the formula:V=QC1

Putting the values for both the capacitors we get:

V1=21.33VV2=26.67V

Therefore, the potential difference isrole="math" localid="1664252368798" 21.33V and26.67V

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