NEET-XII-Physics
39: Alternating Current
- #12In a series RC circuit with an AC source, R = 300 Ω, C = 25 μF, ε0 = 50 V and ν = 50/π Hz. Find the peak current and the average power dissipated in the circuit.Ans : Given:
Resistance of the series RC circuit, R = 300 Ω
Capacitance of the series RC circuit, C = 25 μF
Peak value of voltage, ε0 = 50 V
Frequency of the AC source, ν = 50/`` \,\mathrm{\,\pi \,}`` Hz
Capacitive reactance `` \left({X}_{C}\right)`` is given by,
`` {X}_{C}=\frac{1}{\omega C}``
Here, `` \omega `` = angular frequency of AC source
C = capacitive reactance of capacitance
`` \therefore {X}_{C}=\frac{1}{2\,\mathrm{\,\pi \,}\times {\displaystyle \frac{50}{\,\mathrm{\,\pi \,}}}\times 25\times {10}^{-6}}``
`` \Rightarrow {X}_{C}=\frac{{10}^{4}}{25}\,\mathrm{\,\Omega \,}``
Net reactance of the series RC circuit `` \left(Z\right)`` = `` \sqrt{{R}^{2}+{\left({X}_{C}\right)}^{2}}``
`` \Rightarrow `` Z = `` \sqrt{{\left(300\right)}^{2}+{\left(\frac{{10}^{4}}{25}\right)}^{2}}``
= `` \sqrt{{\left(300\right)}^{2}+{\left(400\right)}^{2}}`` = 500 `` \,\mathrm{\,\Omega \,}``
(a) Peak value of current `` \left({I}_{0}\right)`` is given by,
`` {I}_{0}=\frac{{\epsilon }_{0}}{Z}``
`` \Rightarrow {I}_{0}=\frac{50}{500}=0.1\,\mathrm{\,A\,}``
(b) Average power dissipated in the circuit `` \left(P\right)`` is given by,
P = `` {\epsilon }_{\,\mathrm{\,rms\,}}``Irms cosϕ.
`` {\epsilon }_{rms}`` = `` \frac{{\epsilon }_{0}}{\sqrt{2}}``
and `` {I}_{\,\mathrm{\,rms\,}}=\frac{{I}_{0}}{\sqrt{2}}``
`` \therefore P=\frac{{E}_{0}}{\sqrt{2}}\times \frac{{I}_{0}}{\sqrt{2}}\times \frac{R}{Z}``
`` \Rightarrow P=\frac{50\times 0.1\times 300}{2\times 500}``
`` \Rightarrow P=\frac{3}{2}=1.5\,\mathrm{\,W\,}``
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