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.?

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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.?

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mhz vee

1 week ago

1 week ago

(1) Capacitance= C= 200uF

Potential difference= p.d.= V= 100volts

Work done= W

W= ½(QV)= ½(CV²)

W= ½(200 × 10^-6 × 100²)

= ½(200 × 10^-6 × 10^4)

= ½(200 × 10^-2)

=½(2 × 10^2 × 10^-2)

=½(2 × 10^0)

= ½(2 × 1)

= 1Coulomb volts.

(3) C1= 8uF, C2= 12uF , C3= 24uF ; V= 90volts

(a) C(Effective)= CE= (1/C1 + 1/C2 + 1/C3)^-1 (In series connection)

CE= (1/8 + 1/12 + 1/24)^-1

CE= ( 3 + 2 + 1 /24 )^-1

CE= (6/24)^-1

CE= (1/4)^-1

CE= 4

CE= 4uF

The effective capacitance equals 4uF.

(b) (Effective capacitance / Capacitor) × p.d.= Amount of p.d. the capacitor (For series connection)

At 8uF,

4/8 x 90= 1/2 × 90= 45

Thus, the amount of p.d. in the 8uF capacitor is 45volts.

This means the higher the capacitor value the lower the p.d. stored(From Q= CV, C is inversely proportonal to V).

Potential difference= p.d.= V= 100volts

Work done= W

W= ½(QV)= ½(CV²)

W= ½(200 × 10^-6 × 100²)

= ½(200 × 10^-6 × 10^4)

= ½(200 × 10^-2)

=½(2 × 10^2 × 10^-2)

=½(2 × 10^0)

= ½(2 × 1)

= 1Coulomb volts.

(3) C1= 8uF, C2= 12uF , C3= 24uF ; V= 90volts

(a) C(Effective)= CE= (1/C1 + 1/C2 + 1/C3)^-1 (In series connection)

CE= (1/8 + 1/12 + 1/24)^-1

CE= ( 3 + 2 + 1 /24 )^-1

CE= (6/24)^-1

CE= (1/4)^-1

CE= 4

CE= 4uF

The effective capacitance equals 4uF.

(b) (Effective capacitance / Capacitor) × p.d.= Amount of p.d. the capacitor (For series connection)

At 8uF,

4/8 x 90= 1/2 × 90= 45

Thus, the amount of p.d. in the 8uF capacitor is 45volts.

This means the higher the capacitor value the lower the p.d. stored(From Q= CV, C is inversely proportonal to V).

1 week ago

Potential difference= p.d.= V= 100volts

Work done= W

W= ½(QV)= ½(CV²)

W= ½(200 × 10^-6 × 100²)

= ½(200 × 10^-6 × 10^4)

= ½(200 × 10^-2)

=½(2 × 10^2 × 10^-2)

=½(2 × 10^0)

= ½(2 × 1)

= 1Coulomb volts.

(3) C1= 8uF, C2= 12uF , C3= 24uF ; V= 90volts

(a) C(Effective)= CE= (1/C1 + 1/C2 + 1/C3)^-1 (In series connection)

CE= (1/8 + 1/12 + 1/24)^-1

CE= ( 3 + 2 + 1 /24 )^-1

CE= (6/24)^-1

CE= (1/4)^-1

CE= 4

CE= 4uF

The effective capacitance equals 4uF.

(b) (Effective capacitance / Capacitor) × p.d.= Amount of p.d. the capacitor (For series connection)

At 8uF,

4/8 x 90= 1/2 × 90= 45

Thus, the amount of p.d. in the 8uF capacitor is 45volts.

This means the higher the capacitor value the lower the p.d. stored(From Q= CV, C is inversely proportonal to V).