NEET-XII-Physics
39: Alternating Current
- #4Can the peak voltage across the inductor be greater than the peak voltage of the source in an LCR circuit?Ans : Let a LCR circuit is connected across an AC supply with the emf E = E0 sin ωt.
Let the inductance in the circuit be L
Let the net impedence of the circuit be `` Z=\sqrt{{R}^{2}+({X}_{\,\mathrm{\,L\,}}-{X}_{\,\mathrm{\,C\,}}{)}^{2}}``
Where,
R = resistance in the circuit
XL = reactance due to inductor
XC = reactance due to capacitor
The magnitude of the voltage across the inductor is given by
`` V=L\frac{\,\mathrm{\,d\,}i}{\,\mathrm{\,d\,}t}``
The current in the circuit can be written as `` I={I}_{0}\,\mathrm{\,sin\,}(\omega t+\varphi )``
Where, Ï• is the phase difference between the current and the supply voltage
Thus, the voltage across the inductor can be written as
`` V=L{I}_{0}\,\mathrm{\,cos\,}(\omega t+\varphi )``
Thus peak value of the voltage across the inductor is given by
`` V=L{I}_{0}``
`` \Rightarrow V=\frac{{E}_{0}}{Z}\times L``
Therefore, the peak voltage across the inductor is given by `` V=\frac{{E}_{0}}{Z}\times L``
At resonance Z = R,
`` V=\frac{{E}_{0}}{R}\times L``,
If `` \frac{L}{R}>1``
V > E0
Therefore if magnitude of `` \frac{L}{R}>1`` at resonance the value of the voltage across the inductor will bw greater than the peak value of the supply voltage.
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