1.calculate the work done on a capacitor of capacitance of 200uF, when a potential difference of 100v is applied
2.find the area of a capacitor 1pF,if the distance between the plate is 76cm(=8.85×10^-12f/m)
3.A series arrangement of three capacitors of value 8uf,12uf,24uf is connected in series with a 90v.
a.calculate the effective capacitance in the circuit
b.determine the potential difference across the 8uf capacitor.?
1. To calculate the work done on a capacitor, you can use the formula:
Work = (1/2) * C * V^2
C is the capacitance of the capacitor (in Farads) and V is the potential difference (in Volts).
Plugging in the values given:
Work = (1/2) * (200e-6 F) * (100 V)^2
= (1/2) * (200e-6) * 10000
= 1 J
Therefore, the work done on the capacitor is 1 Joule.
2. The area of a capacitor can be calculated using the formula:
Area = (Capacitance * distance) / constant
Plugging in the values given:
Area = (1e-12 F) * (76 cm) / (8.85e-12 F/m)
= (1e-12) * (76e-2) / (8.85e-12)
= 0.8592 m^2
Therefore, the area of the capacitor is 0.8592 square meters.
3a. To calculate the effective capacitance in a series arrangement, you can use the formula:
1/Ceq = 1/C1 + 1/C2 + 1/C3
Plugging in the values given:
1/Ceq = 1/8e-6 + 1/12e-6 + 1/24e-6
= (3 + 2 + 1)/24e-6
= 6/24e-6
= 250000 F
Ceq = 1 / (250000 F)
= 4e-6 F
Therefore, the effective capacitance in the circuit is 4 microfarads.
3b. To determine the potential difference across the 8uF capacitor, you can use the formula:
V = (Ceq / C3) * Vtotal
Plugging in the values given:
V = (4e-6 F / 8e-6 F) * 90 V
= (0.5) * 90
= 45 V
Therefore, the potential difference across the 8uF capacitor is 45 Volts.
1. To calculate the work done on a capacitor, you can use the formula:
Work = (1/2) * C * V^2
where C is the capacitance and V is the potential difference.
Given:
Capacitance (C) = 200uF = 200 * 10^-6 F
Potential difference (V) = 100V
Substituting the values into the formula:
Work = (1/2) * (200 * 10^-6 F) * (100V)^2
= (1/2) * (200 * 10^-6 F) * (10000 V^2)
= 1 * 10^-4 F * 10,000 V^2
= 1 * 10^-4 * 10,000 V^2
= 1V^2 Joules
Therefore, the work done on the capacitor is 1V^2 Joules.
2. To calculate the area of a capacitor, you can use the formula:
Capacitance (C) = (ε0 * A) / d
where ε0 is the permittivity of free space, A is the area of the capacitor's plates, and d is the distance between the plates.
Given:
Capacitance (C) = 1pF = 1 * 10^-12 F
Distance between plates (d) = 76cm = 76 * 10^-2 m (converting cm to m)
Permittivity of free space (ε0) = 8.85 * 10^-12 F/m
Rearranging the formula to solve for the area (A):
A = C * d / ε0
Substituting the given values:
A = (1 * 10^-12 F) * (76 * 10^-2 m) / (8.85 * 10^-12 F/m)
A = (1 * 10^-12 * 76 * 10^-2) / 8.85 * 10^-12
A ≈ 0.8596 m^2
Therefore, the area of the capacitor is approximately 0.8596 m^2.
3a. To calculate the effective capacitance in a series arrangement of capacitors, you can use the formula:
1/C_effective = 1/C1 + 1/C2 + 1/C3
where C_effective is the effective capacitance, and C1, C2, and C3 are the individual capacitances.
Given:
C1 = 8μF = 8 * 10^-6 F
C2 = 12μF = 12 * 10^-6 F
C3 = 24μF = 24 * 10^-6 F
Substituting the given values into the formula:
1/C_effective = 1/(8 * 10^-6 F) + 1/(12 * 10^-6 F) + 1/(24 * 10^-6 F)
1/C_effective = 125000 + 83333.3333 + 41666.6667
1/C_effective = 249999.9999
Taking the reciprocal of both sides:
C_effective = 1 / (1/C_effective)
C_effective ≈ 4 * 10^-6 F
Therefore, the effective capacitance in the circuit is approximately 4μF.
3b. The potential difference across each capacitor in a series arrangement is the same. Therefore, the potential difference across the 8μF capacitor is the same as the potential difference across the circuit, which is 90V.