Capacitors in Parallel and in Series Problems And Solutions

 Problem#1

Three capacitors C₁ = 0.1μF, C₂ = 0.2μF and C₃ = 0.3μF are connected with 9 V batteries between points A and B. Determine (a) total capacitor capacity, (b) charge and potential difference of each capacitor , and (c) total charge!

Fig.01

Answer:
(a) The total capacity for the series of capacitors arranged in series (for C₁ and C₂) is

1/C₁₂ = 1/C₁ + 1/C₂ = 1/(0.1 μF) + 1/(0.2) μF = 3/0.2 μF

Then C₁₂ = 1/15 μF

C_AB = C₁₂ + C₃ = 1/15 μF + 0.3 μF = 11/30 μF

(b) C₃ capacitors are arranged in parallel with the capacitor (C₁ + C₂) = C₁₂, then the potential difference for capacitor C₃ namely V₃ is equal to the potential difference of the C₁₂ series capacitor namely V₁₂ which is equal to the source potential difference, V₃ = V₁₂ = V = 9 Volt

While the charge on capacitor C₃ is q₃ = C₃V₃ = 0.3μF x 9 V = 2.7 μC

And the charge on C₁₂ series array capacitors is q₁₂ = C₁₂V₁₂ = 1/15 μF x 9 V = 0.6 μC

WATCH THE CONSTRUCTION OF C₁ AND C₂ SERIES

The charge on the capacitor arranged in series is equal to the total charge of the series then

q₁ = q₂ = q₁₂ = 0.6 Μc

The potential difference in both capacitors C1 and C2 arranged in series is

V₁ = q/C₁ = 0.6μC/0.1μF = 6 V

V₂ = q/C₂ = 0.6μC/0.2μF = 3 V

Problem#2
Seven capacitors are arranged as shown in the picture above. If at the end of X and Y a 24 V voltage source is installed, determine: (a) Total capacitor capacity and (b) potential difference between points X and Y.

Fig.02

Answer:
(a) The total capacitor circuit is

Capacitors C₂, C₃ and C₄ are arranged in parallel, then

C₂₃₄ = C₂ + C₃ + C₄ = 1μF + 2μF + 3μF = 6μF

Capacitors C₅ and C₆ are arranged parallel, then

C₅₆ = C₅ + C₆ = 4μF + 5μF = 9μF

Capacitors C₁ and C₂₃₄ are arranged in series, then

1/C₁₂₃₄ = 1/C₁ + 1/C₂₃₄ = 1/(12μF) + 1/(6μF) = 1/(4μF)

C₁₂₃₄ = 4μF

Capacitors C₇ and C₅₆ are arranged in series, then

1/C₅₆₇ = 1/C₅₆ + 1/C₇ = 1/(9μF) + 1/(18μF) = 1/(6μF)

C₅₆₇ = 6μF

Capacitors C₅₆₇ and C₁₂₃₄ are arranged parallel, then C₁₂₃₄ C₅₆₇

C_TOTAL = C₅₆₇ + C₁₂₃₄ = 4μF + 6μF = 10 μF

(b) Note the capacitors C₁₂₃₄ and C₅₆₇ arranged in parallel, then the potential difference between the two capacitors is the same which is equal to the source voltage, then V₁₂₃₄ = V₅₆₇ = V source = 24 V

Problem#3
A capacitor is made of two square metal pieces with area A separated by distance d and has a capacity of C₀ if the two pieces contain air. Determine the capacitor's capacity if half the space between the two pieces is inserted with the dielectric constant K₁ and half the other space is filled with material with its dielectric constant K₂, (a) half the space between the two pieces of material with the dielectric constant K₁ and the other half filled with the dielectric constant K₂ and K₃ as shown in (i), and (b) one third of the space between the two pieces is inserted material with the dielectric constant K₂ and K₃ (figure (ii))

Answer:
(a) capacitors that have a capacity of C₀ after the material has been inserted as shown in (c) into three capacitors which have a capacity of C₁ for materials K₁, C₂ for materials K₂ and C₃ for K₃ materials. Capacitors C₂ and C₃ are arranged in parallel and are series with C₁ with

C₁ = K₁ε₀A/(d/2) = 2K₁ε₀A/d = 2K₁C₀

C₂ = K₂ε₀ (A/2)/(d/2) = K₂C₀ and

C₃ = K₃ε₀ (A / 2) / (d / 2) = K₃C₀

The parallel for C₁ and C₂ is

C₂₃ = C₂ + C₃ = K₂C₀ + K₃C₀

C₂₃ = (K₂ + K₃)C₀

While C₂₃ series with C₁, the capacitor capacity becomes

C_total = C₁C₂₃ /(C₁ + C₂₃)

         = (2K₁C₀)[(K₂ + K₃) C₀]/[(2K₁C₀) + (K₂ + K₃)C₀]

C_total = 2C₀K₁(K₂ + K₃)/[2K₁ + K₂ + K₃]

(b) capacitors that have a capacity of C₀ after being inserted as material (d) into three capacitors that have a capacity of C₁ for air materials that occupy 2/3 of the space, C₂ for materials K2 and C₃ for K3 materials. Capacitors C₂ and C₃ are arranged in parallel and are series with C₁ with

C₁ = ε₀A/(2d/3) = 3ε₀A/2d = 3C₀/2

C₂ = K₂ε₀ (A/2)/(d/3) = 3K₂C₀/2 and

C₃ = K₃ε₀(A/2)/(d/2) = 3K₃C₀/2

The parallel for C₁ and C₂ is

C₂₃ = C₂ + C₃ = 3K₂C₀/2 + 3K₃C₀/2

C₂₃ = 3(K₂ + K₃)C₀/2

While C₂₃ series with C₁, the capacitor capacity becomes

C_total = C₁C₂₃/(C₁ + C₂₃)

        = (3C₀/2) [3(K₂ + K₃) C₀/2]/[(3C₀/2) + 3(K₂ + K₃)C₀/2]

C_total = 3C₀(K₂ + K₃)/2(1 + K₂ + K₃)   

Share :