Question

Four capacitors are connected as shown. Calculate the equivalent capacitance between points P and Q.​

Answer 1

hey look for such questions u can use this trick. firstly mark the endpoints with sm alphabet. then start from a point n till no capacitor or resistance comes mark all the junctions or bends wid the same alphabet with which u started. see i hv done that in the soln. Do the same for the other end too.when u can no longer use the 2 aalphabets use new ones like i did. Finally for the equivalent capacitance or resistance write the alphabets in the order keeping the initial n final end points at terminals n connect all the points by making the resistances or capacitors like i hv done below. u can then easily calculate it. i knw its a bit confusing at first but practice sm no. of similar questions n this method bcms very easy n helpful n saves lot of time.

Answer 2

Calculation of Equivalent Capacitance between two points can be classified on the basis of their connection pattern:

1. For capacitors connected in a series combination, the reciprocal of the equivalent capacitance is the sum of reciprocals of individual capacitances:          $\frac{1}{C_{Eq.}} = \frac{1}{C_{1}} + \frac{1}{C_{2}} + \frac{1}{C_{3}} + .......$
2. For capacitors connected in a parallel combination, the equivalent capacitance is the sum of all individual capacitances in the network:                  $C_{Eq.} = C_{1} + C_{2} + C_{3} + .....$

In the provided Circuit Diagram, we can observe two types of connection between Point P and Point Q:

• Parallel Combination :  2μF, 3μF and 5μF
• Series Combination: [2μF, 3μF, 5μF] and 10μF

Equivalent Capacitance for Parallel Combination, i.e. 2μF, 3μF and 5μF :

2μF + 3μF + 5μF = 10μF

Equivalent Capacitance for Series Combination, i.e. Equivalent for [2μF, 3μF, 5μF] and 10μF  :

$\frac{1}{C_{Eq.}}=\frac{1}{10} + \frac{1}{10}\\\frac{1}{C_{Eq.}}=\frac{2}{10}=\frac{1}{5}$

Therefore,

Equivalent Capacitance for the Given Circuit Diagram is :

5μF

Learn more on Capacitance:

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