Explanation
A
The maximum power transfer theorems states that the maximum power is transferred to a load connected to a capacitive. Same when the load impedance is the complex conjugate of circuit output impedance.
An oscillating circuit contains an inductor of inductance 10−6 H and two capacitor each of capacitance 5×10−6 farad connected in parallel. Then the resonance frequency of the circuit is
When 100\ volt\ DC is applied across a solenoid, a current of 1.0\ amp flows in it. When 100\ volt\ AC is applied across the same coil, the current drops to 0.5\ amp. If the frequency of the AC source is 50\ Hz the impedance and inductance of the solenoid are:
The peak value of an alternating e.m.f. E given by E={ E }_{ 0 } \cos\omega t is 10 volt and frequency is 50 Hz. At time t = (1/600) sec, the instantaneous value of e.m.f. is
A series LCR circuit is tuned to resonance. If the angular frequency of the applied AC voltage at resonance is \omega , the impedance of the circuit then is:
Given,
Resistance, R = 3 \, k\Omega = 3000 \Omega
Capacitance, C = 0.05 \, \mu F = 0.05 \times 10^{-6} \, F
Inductance, L = 120 \, mH = 120 \times 10^{-3} \, H
f = 5 \, kHz = 5000 \, Hz
It is asked to find the impedance in polar form
We have ,
Impedence, Z = \sqrt{R^2 + (X_L – X_C)^2}
Also
Inductive Reactance, X_L = L \omega , X_C = \dfrac{1}{C \omega}
We have \omega = 2 \pi F = 2 \times 3.14 \times 5000 = 31416
X_L = L \omega = 120 \times 10^{-3} \times 31416 = 3769.9
Capacitive Reactance, X_C = \dfrac{1}{C \omega} = \dfrac{1}{31416 \times 0.05 \times 10^{-6}} = 636.61
Z = \sqrt{3000^2 + (3769.9 – 636.61)^2} = \sqrt{9000,000 + 9819090.9} = 4337.2 \approx 4337 \Omega
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