The tuning circuit of a radio receiver has a resistance of 50Ω, an inductor of 10 mH and a variable capacitor. AA1 MHz radio wave produces a potential difference of 0.1 mV. The values of the capacitor to produce resonance is(Take π2 = 10)
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In a circuit inductance L and capacitance C are connected as shown in figure.A1 and A2 are ammeters.
When key K is pressed to complete the circuit, then just after closing key (K), the readings of A1 and A2 will be
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Initially there is no D.C. current in inductive circuit and maximum D.C.current is in capacitive current. Hence,the current is zero in A2 and maximum in A1.
A capacitor in an ideal LC circuit is fully charged by a DC source, then it is disconnected from DC source, the current in the circuit
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In an LR circuit f = 50 Hz, L=2H, E=5 volts, R=1 Ω then energy stored in inductor is
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In an oscillating LC circuit the max. charge on the capacitor is Q. The charge on capacitor when the energy is stored equally between electric and magnetic field is
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For the circuit shown in the fig., the current through the inductor is 0.9 A while the current through the condenser is 0.4A. Then
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The current drawn by inductor and capacitor will be in opposite phase.Hence net current drawn from generator = IL– IC= 0.9 – 0.4 = 0.5 amp.
In an A.C. circuit,a resistance of R ohm is connected in series with an inductance L. If phase angle between voltage and current be 45°, the value of inductive reactance will be
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