NEET-XII-Physics
39: Alternating Current
- #2An AC source producing emf
ε = ε0 [cos (100 π s-1)t + cos (500 π s-1)t]
is connected in series with a capacitor and a resistor. The steady-state current in the circuit is found to be i = i1 cos [(100 π s-1)t + φ1) + i2 cos [(500π s-1)t + Ï•2]. So,
(a) i1 > i2
(b) i1 = i2
(c) i1 < i2
(d) The information is insufficient to find the relation between i1 and i2.digAnsr: cAns : (c) i1 < i2
The charge on the capacitor during steady state is given by,
`` Q=C\epsilon ={\epsilon }_{0}C\left[\,\mathrm{\,cos\,}\left(100{\,\mathrm{\,\pi s\,}}^{-1}\right)t+\,\mathrm{\,cos\,}\left(500{\,\mathrm{\,\pi s\,}}^{-1}\right)t\right]``
The steady state current is, thus, given by,
`` i=\frac{dQ}{dt}={\epsilon }_{0}C\times 100\,\mathrm{\,\pi \,}\left[\,\mathrm{\,sin\,}\left(100{\,\mathrm{\,\pi s\,}}^{-1}\right)t\right]+{\epsilon }_{\mathit{0}}C\times 500\,\mathrm{\,\pi \,}\left[\,\mathrm{\,sin\,}\left(500{\,\mathrm{\,\pi s\,}}^{-1}\right)t\right]``
`` \Rightarrow i=100C{\,\mathrm{\,\pi \epsilon \,}}_{0}\,\mathrm{\,cos\,}\left[\left(100{\,\mathrm{\,\pi s\,}}^{-1}\right)\,\mathrm{\,t\,}+{\varphi }_{1}\right]+500C{\,\mathrm{\,\pi \epsilon \,}}_{0}\,\mathrm{\,cos\,}\left[\left(500{\,\mathrm{\,\pi s\,}}^{-1}\right)+{\varphi }_{2}\right]``
`` \Rightarrow {i}_{1}=100C{\,\mathrm{\,\pi \epsilon \,}}_{0}\&{i}_{\mathit{2}}=500C{\,\mathrm{\,\pi \epsilon \,}}_{0}``
`` \therefore {i}_{2}>{i}_{1}``
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