Capacitors

JEE Main level, numerical answer, medium

159 of 180

Question 159 of 180

A capacitor of capacitance 900µF900µF is charged by a 100V100V battery. The capacitor is disconnected from the battery and connected to another uncharged identical capacitor such that one plate of uncharged capacitor connected to positive plate and another plate of uncharged capacitor connected to negative plate of the charged capacitor. The loss of energy in this process is measured as x×10−2Jx\times {10}^{-2}J. The value of xx is ______.

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Answer: 225

When a charged capacitor is connected to an uncharged capacitor, then there is loss of energy in the form of heat as charges flow from higher potential to lower potential.
So, heat ∆H=12C1C2C1+C2(V1−V2)2∆H=\dfrac{1}{2}\dfrac{{C}_{1}{C}_{2}}{{C}_{1}+{C}_{2}}{\left({V}_{1}-{V}_{2}\right)}^{2}
As the other capacitor is identical therefore charge is equally divided.
=12C22C(100−0)2=\dfrac{1}{2}\dfrac{{C}^{2}}{2C}(100-0{)}^{2}
=12900×10−62×104=94=\dfrac{1}{2}\dfrac{900\times {10}^{-6}}{2}\times {10}^{4}=\dfrac{9}{4} Joule
=2.25=2.25 Joule.
Hence, the value of x=225x=225.

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