Question

# Two condensers are in parallel and the energy of the combination is 0.1 J, when the difference of potential between terminals is 2 V. With the same two condensers in series, the energy is $$1.6\times 10^{-2}$$J for the same difference of potential across the series combination. What are the capacities?

Solution

## Let the capacities are $$C_1, C_2$$When they are in parallel, the net capacitance $$C_p=C_1+C_2$$Thus, energy stored for this combination, $$U_p=\dfrac{1}{2}C_pV^2$$or $$0.1=0.5(C_1+C_2)(2^2)$$ or $$C_1+C_2=0.05 ...(1)$$When they are in series, the net capacitance $$C_s=\dfrac{C_1C_2}{C_1+C_2}$$Thus, energy stored for this combination, $$U_s=\dfrac{1}{2}C_sV^2$$or $$1.6\times 10^{-2}=0.5\dfrac{C_1C_2}{C_1+C_2}(2^2)$$ or $$C_1+C_2=125C_1C_2 ...(2)$$From (1) and (2),  $$C_1+(0.05-C_1)=125C_1(0.05-C_1)$$or $$0.05=6.25C_1-125C_1^2$$or $$1=125C_1-2500C_1^2$$ or $$2500C_1^2-125C_1+1=0$$so, $$C_1=\dfrac{125\pm\sqrt{125^2-4(2500)(1)}}{2(2500)}=\dfrac{125\pm 75}{5000}$$So, $$C_1=40 mJ$$  or $$10 mJ$$Thus, from (1) $$C_2=10 mJ$$ or $$40 mJ$$PhysicsNCERTStandard XII

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