  Question

The circuit involves two ideal cells connected to a 1μF capacitor via a key K. Initially the key K is in  position 1 and the capacitor is charged fully by 2 V cell. The key is pushed to position 2. Column I gives physical quantities involving the circuit after the key is pushed from position 1 to position 2. Column II gives corresponding results. Match the statements in Column I with the corresponding values in column II. Column IColumn II(A) The net charge crossing the 4 volt cell in μC is       (p)  2(B) The magnitude of work done by 4 volt cell in μJ is (q)  6(C) The gain in potential energy of capacitor in μJ is  (r)  8(D) The net heat produced in μJ is                                (s)  16

A
Ap,Bq:Cr:Ds  B
Ap:Br:Cq:Dp  C
Aq:Bp:Cr:Ds  D
As:Bs:Cp:Dp  Solution

The correct option is B A→p:B→r:C→q:D→pWhen connected to 1, charge on capacitor Qi=CVi=2 μC(∵Vi=2V)Energy stored in the capacitor Ui=12C(Vi)2=2 μJ When key is shifted to 2 (Polarity remains same) Qf=CVf=4 μC(∵Vf=4V)Uf=12CV2f=8 μJNet charge crossing 4 volt cellΔQ=Qf−Qi=4−2=2 μCGain in energy stored in the capacitor ΔU=Uf−UiΔU=8−2ΔU=6 μJWork done by 4 Volt cellW4V=△QVfW4V=(Qf−Qi)VfW4V=(4−2)4  (∵△Q=2μC)W4V=8μJ Also from conservation of energy , Ui+W4V=Uf+H,Here H is heat dissipated. ⇒2+8=8+H⇒H=2 μJPhysics

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