A 1 m long conducting rod rotates with an angular frequency of 400 rad \[{{s}^{-1}}\] about an axis normal to the rod passing through its one end. |
The other end of the rod is in contact with a circular metallic ring. A constant magnetic field of 0.5 T parallel to the axis exists everywhere, |
Find an expression for the emf induced between the centre and the ring and then calculate the value of induced emf from the given data, |
Or |
A circuit containing a 80 m H inductor and a \[60\,\mu F\] capacitor in series is connected to a 230 V 50 Hz supply. The resistance of the circuit is negligible. |
(i) Obtain the current amplitude and rms values. |
(ii) Obtain the rms values of potential drops across each element. |
(iii) What is the average power transferred to the inductor? |
(iv) What is the average power transferred to the capacitor? |
(v) What is the average power absorbed by the circuit? |
Answer:
Emf induced between the centre and the ring \[e=\frac{d\phi }{dt}=\frac{d(BA)}{dt}\] \[[\because \phi =BA]\] \[e=B\times \frac{dA}{dt}\] ?(i) Let the rod subtends an angle, \[\theta \] at centre in time t, then area covered by it \[A=\frac{1}{2}{{R}^{2}}\theta \] \[\therefore \frac{dA}{dt}=\frac{1}{2}{{R}^{2}}\frac{d\theta }{dt}\Rightarrow \frac{dA}{dt}=\frac{1}{2}{{R}^{2}}\omega \] where, \[\omega =d\theta /dt\] is the angular velocity. \[\therefore \] Emf induced, \[e=\frac{1}{2}{{R}^{2}}\omega B\] Given, angular frequency, \[\omega =400\,\,rad\,{{s}^{-1}}\] Magnetic field, B = 0.5 T Length of wire, R = L = 1 m Induced emf, \[e=(1/2)\times {{(1)}^{2}}\times 400\times 0.5=100\,V\] Or Given, inductance of inductor, \[L=80\,mH=80\times {{10}^{-3}}H\] Capacitance of capacitor, \[C=60\mu F={{6010}^{-6}}F\] Voltage, V = 230 V, Frequency, v = 50 Hz Impedance of the circuit, \[Z=\left( L\omega -\frac{1}{C\omega } \right)\] or \[Z=\left| L\omega -\frac{1}{C\omega } \right|\] \[=\left| \left( 2\pi \times 50\times 8\times {{10}^{-3}}-\frac{1}{2\pi \times 50\times 60\times {{10}^{-6}}} \right) \right|\] \[=\left| (25.12-53.08) \right|=27.93\,\Omega \] (i) Current, \[{{I}_{rms}}=\frac{V}{Z}=\frac{230}{27.96}A=8.23A\] Current amplitude, \[{{I}_{0}}={{I}_{rms}}\sqrt{2}=8.23\times 1.414A=11.63\,A\] (ii) Potential drop across inductor, \[{{V}_{L}}=I{{X}_{L}}=I\omega L=8.23\times 25.12=206.7\,V\] Potential drop across capacitor, \[{{V}_{C}}=I{{X}_{C}}=I\times \frac{1}{\omega C}=8.23\times 53.08=436.8\,V\] (iii) Average power transferred to the inductor, \[\overline{P}={{V}_{L}}{{I}_{L}}\cos 90{}^\circ =0\] (iv) Average power transferred to the capacitor, \[\overline{P}={{V}_{C}}{{I}_{C}}\cos 90{}^\circ =0\] (v) Total average power absorbed by the circuit is zero.
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