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 $$5\times { 10 }^{ -6 }$$ 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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