NEET-XII-Physics
39: Alternating Current
- #8A capacitor of capacitance 10 μF is connected to an oscillator with output voltage ε = (10 V) sin ωt. Find the peak currents in the circuit for ω = 10 s-1, 100 s-1, 500 s-1 and 1000 s-1.Ans : Capacitance of the capacitor, C = 10 μF = 10 × 10-6 F = 10-5 F
Output voltage of the oscillator, `` \epsilon ``= (10 V)sinωt
On comparing the output voltage of the oscillator with `` \epsilon ={\epsilon }_{0}\,\mathrm{\,sin\,}\omega t``, we get:
Peak voltage `` {\epsilon }_{0}`` = 10 V
For a capacitive circuit,
Reactance, `` {X}_{c}=\frac{1}{\omega C}``
Here, `` \omega `` = angular frequency
C = capacitor of capacitance
Peak current, `` {I}_{0}`` = `` \frac{{\epsilon }_{0}}{{X}_{c}}``
(a) At ω = 10 s-1:
Peak current,
I0 = `` \frac{{\epsilon }_{0}}{{X}_{c}}``
`` =\frac{{\epsilon }_{0}}{1\mathit{/}\omega C}``
`` =\frac{10}{1/10\times {10}^{-5}}\,\mathrm{\,A\,}``
= 1 × 10-3 A
(b) At ω = 100 s-1:
Peak current, I0 = `` \frac{{\epsilon }_{0}}{1/\omega C}``
`` \Rightarrow {I}_{0}=\frac{10}{1/100\times {10}^{-5}}``
`` \Rightarrow {I}_{0}=\frac{10}{{10}^{3}}=1\times {10}^{-2}\,\mathrm{\,A\,}``
`` =0.01\,\mathrm{\,A\,}``
(c) At ω = 500 s-1:
Peak current, I0 = `` \frac{{\epsilon }_{0}}{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\omega C$}\right.}}``
`` {I}_{0}=\frac{{\epsilon }_{0}}{1/\omega C}``
`` \Rightarrow {I}_{0}=\frac{10}{1/500\times {10}^{-5}}``
`` \Rightarrow {I}_{0}=10\times 500\times {10}^{-5}``
`` =5\times {10}^{-2}\,\mathrm{\,A\,}=0.05\,\mathrm{\,A\,}``
(d) At ω = 1000 s-1:
Peak current, I0 = `` \frac{{\epsilon }_{0}}{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\omega C$}\right.}}``
`` \Rightarrow {I}_{0}=\frac{10}{1/1000\times {10}^{-5}}``
`` \Rightarrow {I}_{0}=10\times 1000\times {10}^{-5}``
`` \Rightarrow {I}_{0}={10}^{-1}\,\mathrm{\,A\,}=0.1\,\mathrm{\,A\,}``
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