NEET-XII-Physics
39: Alternating Current
- #15Consider the situation of the previous problem. Find the average electric field energy stored in the capacitor and the average magnetic field energy stored in the coil.Ans : Given:
Resistance in the LCR circuit, R = 300 Ω
Capacitance in the LCR circuit, C = 20 μF = 20 × 10-6 F
Inductance in the LCR circuit, L = 1 henry
Net impedance of the LCR circuit, Z = 500 ohm
RMS value of voltage, `` {\epsilon }_{\,\mathrm{\,rms\,}}`` = 50 V
RMS value of current, Irms = 0.1 A
Peak current `` \left({I}_{0}\right)`` is given by,
`` {I}_{0}=\frac{{E}_{\,\mathrm{\,rms\,}}\sqrt{\mathit{2}}}{Z}=\frac{50\times 1.414}{500}=0.1414\,\mathrm{\,A\,}``
Electrical energy stored in capacitor`` \left({U}_{C}\right)`` is given by,
`` {U}_{C}=\frac{1}{2}C{V}^{2}``
`` \Rightarrow {U}_{C}=\frac{1}{2}\times 20\times {10}^{-6}\times 50\times 50``
`` \Rightarrow {U}_{C}=25\times {10}^{-3}\,\mathrm{\,J\,}=25\,\mathrm{\,mJ\,}``
Magnetic field energy stored in the coil `` \left({U}_{L}\right)`` is given by,
`` {U}_{L}=\frac{1}{2}L{I}_{0}^{2}``
`` \Rightarrow {U}_{L}=\frac{1}{2}\times 1\times {\left(0.1414\right)}^{2}``
`` \Rightarrow {U}_{L}\approx 5\times {10}^{-3}\,\mathrm{\,J\,}``
`` \Rightarrow {U}_{L}=5\,\mathrm{\,mJ\,}``
Page No 330: