Solveeit Logo
Login

Question

Question: In the figure \({{C}_{1}}=10.0\mu F\), \({{C}_{2}}=20.0\mu F\)and \({{C}_{3}}=25.0\mu F\). If no cap...

In the figure C1=10.0μF{{C}_{1}}=10.0\mu F, C2=20.0μF{{C}_{2}}=20.0\mu Fand C3=25.0μF{{C}_{3}}=25.0\mu F. If no capacitor can withstand a potential difference of more than 100V without failure what are:
(a) The magnitude of the maximum potential difference that can exist between points A and B and
(b) the maximum energy that can be stored in the three-capacitor arrangement?

Explanation

Solution

For the first part, find the potential difference across each capacitor and then find their sum to find the net potential difference across the given terminals. Now, for the second part, you could find the effective capacitance of the given system, then substitute this in the formula and hence find the energy.

Complete step-by-step solution:
In the question, we are given an arrangement of three capacitors of capacitances: C1=10.0μF{{C}_{1}}=10.0\mu F, C2=20.0μF{{C}_{2}}=20.0\mu Fand C3=25.0μF{{C}_{3}}=25.0\mu F. We are also told that none of these capacitors can withstand potential differences that are higher than 100V.
We are supposed to find the magnitude of maximum potential difference that can exist between A and B. Also, the maximum energy that could be stored in this arrangement should be found.
(a) We know that,
Q=CVQ=CV
With the same charge stored in every capacitor, the maximum voltage would be present across the capacitor of least capacitance that is, C1{{C}_{1}}.
With this knowledge, we could say that:
Voltage across C2{{C}_{2}}of 20μF20\mu Fcapacitance would be 50V and that across C3{{C}_{3}}of 25μF25\mu Fcapacitance would be 40V.
So, the net voltage across A and B would be the sum here. That would be 190V.
(b) Energy stored in a capacitor could be given by,
E=12CV2E=\dfrac{1}{2}C{{V}^{2}}
Let us find the effective capacitance,
1C=1C1+1C2+1C3=110μF+120μF+125μF\dfrac{1}{C}=\dfrac{1}{{{C}_{1}}}+\dfrac{1}{{{C}_{2}}}+\dfrac{1}{{{C}_{3}}}=\dfrac{1}{10\mu F}+\dfrac{1}{20\mu F}+\dfrac{1}{25\mu F}
⇒C=5.26μF\Rightarrow C=5.26\mu F
So, energy stored in the given system would be,
E=12×5.26×10−6×(190)2=0.095JE=\dfrac{1}{2}\times 5.26\times {{10}^{-6}}\times {{\left( 190 \right)}^{2}}=0.095J
We found the energy stored to be 0.095J.

Note: The formula used for finding the effective capacitance is exclusive for series connection. For the capacitors connected parallel with each other, the net capacitance is found by the sum of the individual capacitances. Mathematically, this could be,
Ceff=C1+C2+C3{{C}_{eff}}={{C}_{1}}+{{C}_{2}}+{{C}_{3}}.