Electromagnetism - SS2 Physics Past Questions and Answers - page 3

Given: C = 5 μF = 5 × 10(-6) F, V = 10 V

Q = C × V

Q = (5 × 10(-6) F) × (10 V)

Q = 50 × 10(-6) C

Q = 50 μC

Therefore, the charge stored on the capacitor is 50 μC.

Given: C = 8 pF = 8 × 10(-12) F, V = 12 V

The energy stored in the capacitor is given by the formula: E = 1/2CV2

E = 1/2 × (8 × 10(-12) F) × (12 V)2

E = 1/2 × 8 × 10(-12) F × 144 V2

E = 576 × 10(-12) F × V2

E = 576 × 10(-12) J

Therefore, the energy stored in the capacitor is 576 × 10(-12) J.

The total capacitance of capacitors connected in parallel is given by the sum of the individual capacitances.

Total capacitance = C1 + C2

Total capacitance = 6 μF + 8 μF

Total capacitance = 14 μF

Therefore, the total capacitance of the combination is 14 μF.

The total capacitance of capacitors connected in series is given by the reciprocal of the sum of the reciprocals of the individual capacitances.

1/Total capacitance = 1/C1 + 1/C2 + 1/C3

1/Total capacitance = 1/4 μF + 1/6 μF + 1/8 μF

1/Total capacitance = (3/12 + 2/12 + 3/24) μF

1/Total capacitance = 8/24 μF

1/Total capacitance = 1/3 μF

Total capacitance = 3 μF

Therefore, the total capacitance of the combination is 3 μF.

Explanation: The time constant (τ) of an RC circuit is given by the product of resistance (R) and capacitance (C). τ = RC = (10,000 Ω) × (100 × 10(-6) F) = 1 × 10(-3) s = 10 ms.