A resistor of 2 Omega,an inductor of 100 mu H | vedclass

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Question

(A) At resonance,the inductive reactance equals the capacitive reactance $(X_L = X_C)$.Given: $R = 2 \Omega$,$L = 100 	imes 10^{-6} 	ext{ H}$,$C = 4

Vedclass · Accepted Answer

$25$

Vedclass · Answer

(A) At resonance,the inductive reactance equals the capacitive reactance $(X_L = X_C)$.Given: $R = 2 \Omega$,$L = 100 	imes 10^{-6} 	ext{ H}$,$C = 400 	imes 10^{-12} 	ext{ F}$,$e_{rms} = 0.1 	ext{ V}$.The impedance at resonance is $Z = R = 2 \Omega$.The current in the circuit is $I_{rms} = \frac{e_{rms}}{Z} = \frac{0.1}{2} = 0.05 	ext{ A}$.The resonant angular frequency is $\omega = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{100 	imes 10^{-6} 	imes 400 	imes 10^{-12}}} = \frac{1}{\sqrt{4 	imes 10^{-14}}} = \frac{1}{2 	imes 10^{-7}} = 0.5 	imes 10^7 = 5 	imes 10^6 	ext{ rad/s}$.The voltage drop across the inductor is $V_L = I_{rms} 	imes X_L = I_{rms} 	imes L\omega$.$V_L = 0.05 	imes (100 	imes 10^{-6}) 	imes (5 	imes 10^6) = 0.05 	imes 10^{-4} 	imes 5 	imes 10^6 = 0.05 	imes 500 = 25 	ext{ V}$.

$A$ resistor of $2 \Omega$,an inductor of $100 \mu H$,and a capacitor of $400 pF$ are connected in series across a source of $e_{rms} = 0.1 \text{ V}$. At resonance,the voltage drop across the inductor is: (in $\text{ V}$)

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