A capacitor of capacitance $$C=3$$ μF is first charged by connecting it across a $$10V$$ battery by closing key $$K1$$. It is then allowed to get discharged through $$2$$ Ω and $$4$$ Ω resistor by closing the key $$K2$$ and opening key $$K1$$ .The total energy dissipated in the $$2$$ Ω resistor is equal to

Question

A capacitor of capacitance  $$C=3$$ μF is first charged by connecting it across a  $$10V$$ battery by closing key $$K1$$. It is then allowed to get discharged through  $$2$$ Ω and  $$4$$ Ω resistor by closing the key $$K2$$ and opening key $$K1$$ .The total energy dissipated in the  $$2$$ Ω resistor is equal to

Infinity Learn · Accepted Answer

Before $$K1$$ opened the energy stored over capacity  $$C1=12CV2=12$$×$$3$$×$$10$$−$$6$$×$$102=150$$ μJNow this energy will be dissipated  $$E=i2Rt$$ in the two resistors in  the ratio of $$R1$$:$$R2$$ as they are in series (current$$ flowing through the resistors is $$same) when switch $$K2$$ is closed.∴$$150$$ μJ is lost at the ratio  $$2$$:$$4=1$$:$$2In$$ $$2$$ Ω , $$13rd$$ of $$150$$ μJ $$=$$ $$50$$ μJ and In $$4$$ Ω ,$$23rd$$ of $$150$$ μ$$J=100$$ μJ will be lost.Therefore, the correct answer is $$($$ B $$)$$

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A capacitor of capacitance C=3 μF is first charged by connecting it across a 10V battery by closing key K1. It is then allowed to get discharged through 2 Ω and 4 Ω resistor by closing the key K2 and opening key K1 .The total energy dissipated in the 2 Ω resistor is equal to

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