Questions
A capacitor of capacitance is first charged by connecting it across a 10V battery by closing key . It is then allowed to get discharged through and resistor by closing the key and opening key .The total energy dissipated in the resistor is equal to
detailed solution
Correct option is B
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 μJ=100 μJ will be lost.Therefore, the correct answer is ( B )Similar Questions
A 600 pF capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. What is the common potential (in V) and energy lost (in J) after reconnection?
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