1.Consider the figure below. (a) Find the charge stored on each capacitor in the

Question

1.Consider the figure below.

(a) Find the charge stored on each capacitor in the figure shown above (C1 = 17.1 µF, C2 = 8.62 µF) when a 1.68 V battery is connected to the combination.

(b) What energy is stored in each capacitor?

Q1 =
What is the total charge stored in the equivalent capacitor? What are the properties of series and parallel combination of capacitors regarding charge and potential difference? C Q2 =
What is the total charge stored in the equivalent capacitor? What are the properties of series and parallel combination of capacitors regarding charge and potential difference? C Q3 =   C

Explanation / Answer

(a)

The capacitors C1 and C2 are connected in parallel. So, the capacitance of the parallel

combination of the two capacitors is,

       C12 = C1 + C2 = 17.1X10-6 F + 8.62x10-6 F = 25.72x 10-6 F

This equivalent capacitance is connected in series with 0.300 µF. Thus, the total capacitance of

the circuit is,

       C = C3C12/C3+C12 = (0.3x10-6 F)(25.72X10-6 F) / [0.3x10-6 F + 25.72x10-6 F] = 0.2965x10-6 F

The net charge on the capacitor is,

     Q = CV = (0.2965x10-6 F)(1.68 V) = 0.49812x10-6 C

Voltage across C12 is,

     V12 = Q/C12 = 0.49812x10-6 C /25.72x 10-6 F = 0.019367 V

The charge Q1 is,

       Q1 = C1V12 = (17.1x10-6 F)(0.019367 V) = 0.33X10-6 C = 3.3x10-7 C

The charge Q2 is,

      Q2 = C2V12 = (8.62x10-6 F)(0.019367 V) = 0.167X10-6 C = 1.67x10-7 C

The charge Q3 is,

       Q3 = Q = 0.49812x10-6 C = 4.98x10-7 C

In parallel combination, the potential difference across the capacitors is the same, but the charge is different.

in series combination, the potential difference across the capacitors is the different, but the charge is the same.

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(b)

Energy stored in C1 is,

     E1 = 0.5[Q12 /C1] = (0.5)(0.33X10-6 C)2 / (17.1X10-6 F) = 3.18x10-9 J = 3.18 nJ

Energy stored in C2 is,

     E2 = 0.5[Q22 /C2] = (0.5)(0.167X10-6 C)2 / (8.62X10-6 F) = 1.617x10-9 J = 1.617nJ

Energy stored in C2 is,

     E3 = 0.5[Q32 /C3] = (0.5)(0.49812x10-6 C )2 / (0.3X10-6 F) = 4.135x10-7 J = 0.4135 µJ

       

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