QUESTION 4 A 25-uF capacitor charged to 50 V and a capacitor C charged to 20 V a

Question

QUESTION 4 A 25-uF capacitor charged to 50 V and a capacitor C charged to 20 V are connected to each other, with the two positive plates connected and the two negative plates connected. The final potential difference across the 25-uF capacitor is 36 V. What is the value of the capacitance of C? O a. C. 22 uF d, 43 F e. 29 uF QUESTION 5 What is the total energy stored in the group of capacitors shown if the charge on the 30-uF capacitor is 0.90 mC? 30 F 20 uF 15 uF a. 32 mJ Ob. 21 mJ c. 29 mJ e. 66 mJ

Explanation / Answer

5) Let’s determine the energy of the 30-µF capacitor.  
Energy = Q^2 ÷ (2 * C) = (0.9 * 10^-3)^2 ÷ 60 * 10^-6 = 1.35 * 10^-2 J

Let’s determine the voltage of the 30-µF capacitor.
C = Q ÷ V, V = Q ÷ C
V = 0.9 * 10^-3 ÷ 30 * 10^-6 = 30 volts

Since the 15 F capacitor is in parallel with the 30 F capacitor, the voltage is the same for both capacitors.
Energy = ½ * C * V^2 = ½ * 15 * 10^-6 * 30^2 = 6.375 * 10^-3

When capacitors are in series, the charge is the same. Let’s determine the charge on the 15 F capacitor.
C = Q ÷ V, Q = C * V
Q = 15 * 10^-6 * 30 = 4.5 * 10^-4 C
Total charge of parallel capacitors = 0.9 * 10^-3 + 4.5 * 10^-4 = 1.35 * 10^-3
Since the 20 F capacitor is in series with the other capacitors, the charge of it is 1.35 * 10^-3.
Energy = Q^2 ÷ (2 * C) = (1.35 * 10^-3)^2 ÷ 40 * 10^-6 = 4.55625 * 10^-2

Total energy = 1.35 * 10^-2 + 6.375 * 10^-3 + 4.55625 * 10^-2 = 6.54375 * 10^-2 J
The total energy is 0.065375 joules

This rounds to 66 mJ

ANS:- e

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