Physics, asked by shaikhayesha4311, 5 months ago

Three capacitors of capacitance 2pF, 3pF and 6pF are connected in series. The equivalent
capacitance of the combination is:

Answers

Answered by rakshahema2003
5

Answer:

Explanation:

Cs= 0.66pF

Q= 66×10-12C

V1= 33V

V2= 22V

V3= 16.5 V

Answered by archanajhaasl
2

Answer:

The equivalent capacitance of the combination is 1pF.

Explanation:

When the capacitors are connected in series the equivalent capacitance is given as,

\frac{1}{C}= \frac{1}{C_1} +\frac{1}{C_2} +\frac{1}{C_3}       (1)

Where,

C=equivalent capacitance of the capacitors connected in series

C₁, C₂, C₃=respective capacitance of the individual capacitors

From the question we have,

C₁=2pF

C₂=3pF

C₃=6pF

By substituting the required values in equation (1) we get;

\frac{1}{C}= \frac{1}{2} +\frac{1}{3} +\frac{1}{6}

\frac{1}{C}=\frac{3+2+1}{6}

\frac{1}{C}=\frac{6}{6}

C=1pF

Hence, the equivalent capacitance of the combination is 1pF.

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