(8c25p65&66) A capacitance C1 = 7.6 F is connected in series with a capacitance

Question

(8c25p65&66) A capacitance C1 = 7.6 F is connected in series with a capacitance C2 = 5.0 F, and a potential difference of 250 V is applied across the pair. Calculate the equivalent capacitance. What is the charge on C1? Tries 0/5 What is the charge on C2? What is the potential difference across C1? What is the potential difference across C2? Repeat for the same two capacitors but with them now connected in parallel. Calculate the equivalent capacitance. What is the charge on C1? What is the charge on C2? What is the potential difference across C1? What is the potential difference across C2?

Explanation / Answer

Here ,

C1 = 7.6 uF

C2 = 5 uF

A) for the capacitors in series

1/Ceq =1/C1 + 1/C2

1/Ceq = 1/7.6 + 1/5

Ceq = 3.06 uF

the equivalent capacitance is 3.06 uF

B)

charge on C1 = Ceq * V

charge on C1 = 3.06 * 250 uC

charge on C1 =754 uC

C)

charge on C2 = Ceq * V

charge on C2 = 3.06 * 250

charge on C2 = 754 uC

D)

potential difference across C1 = Q1/C1

potential difference across C1 = 754/7.6 V

potential difference across C1 = 99.2 V

E)

Potential difference across C2 = 250 - V1

Potential difference across C2 = 250 - 99.2 V

Potential difference across C2 = 150.8 V

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