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CBSE Class 12 Physics 2013 Paper

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Question : 19 of 29
Marks: +1, -0
(a) When an a.c. source is connected to an ideal capacitor show that the average power supplied by the source over a complete cycle is zero.
(b) A lamp is connected in series with a capacitor. Predict your observations when the system is connected first across a d.c. and then an a.c. source. What happens in each case if the capacitance of the capacitor is reduced?
Solution:  
(a) AC source connected to a ideal capacitor:
Using KVL,
  V0sin⁡  ωt=  qC\; V_0 \sin \; \omega t = \; \frac{q}{C}
     Or,       q=CV0sin⁡  ωt\; \; \text{ Or, } \; \; \; q = C V_0 \sin \; \omega t
Or,   I=  dqdt=CV0ωcos⁡ωt\; I = \; \frac{dq}{dt} = C V_0 \omega \cos \omega t
  ∴    I=I0cos⁡ωt=I0sin⁡  (ωt+  π2)\; \therefore \; \; I = I_0 \cos \omega t = I_0 \sin \; \left( \omega t + \; \frac{\pi}{2} \right)
Average Power =  1T∫0T[V0sin⁡  ωt×I0sin⁡  (ωt+π2)]dt= \; \frac{1}{T} \int_{0}^{T} \left[ V_0 \sin \; \omega t \times I_0 \sin \; \left( \omega t + \frac{\pi}{2} \right) \right] dt
  =  1T∫0T[12V0I0sin⁡  2ωt]dt\; = \; \frac{1}{T} \int_{0}^{T} \left[ \frac{1}{2} V_0 I_0 \sin \; 2 \omega t \right] dt
  =0\; = 0
(b) Reactance of a capacitor =XC=  12πfC= X_C = \; \frac{1}{2} \pi f C
For DC, f=0,XC=∞f = 0, X_C = \infty, No current flow, bulb does not glow.
For AC,f≠0,XC≠∞\mathrm{AC}, f \neq 0, X_C \neq \infty, current flows, bulb glows.
Capacitance is reduced.
For DC, f=0,XC=∞f = 0, X_C = \infty, No current flow, bulb does not glow.
For AC\mathrm{AC}, as C\mathrm{C} reduces, XCX_C increases, current decreases, bulb glows less bright.
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