Question 405: AC fundamentals

Question

A series L-C-R circuit containing a resistance of $$120 ~\Omega$$ has angular frequency $$4 	imes 10^5 \mathrm{~rad} \mathrm{~s}^{-1}$$. At resonance the voltage across resistance and inductor are $$60 \mathrm{~V}$$ and $$40 \mathrm{~V}$$ respectively, then the value of inductance will be

Accepted Answer

To find the inductance value in the L-C-R circuit at resonance, we can use the given information about the angular frequency ($$\omega$$), the voltage across the resistance, and the voltage across the inductor at resonance.
The given parameters are:
Resistance, $$R = 120 ~\Omega$$
Angular frequency, $$\omega = 4 	imes 10^5 ~	ext{rad} ~	ext{s}^{-1}$$
Voltage across resistance, $$V_R = 60 ~	ext{V}$$
Voltage across inductor, $$V_L = 40 ~	ext{V}$$
At resonance in an L-C-R circuit, the impedance is purely resistive, i.e., the impedance of the circuit is equal to the resistance $$R$$.
The voltage across the inductor $$V_L$$ at resonance is given by the equation:
$$ V_L = \omega L I $$
where $$I$$ is the current in the circuit, which can be found using Ohm's law considering the voltage across the resistance:
$$ V_R = I \cdot R $$
Rearrange to solve for $$I$$:
$$ I = \frac{V_R}{R} $$
Substitute the given values:
$$ I = \frac{60 ~	ext{V}}{120 ~\Omega} = 0.5 ~	ext{A} $$
Now, use the value of current to solve for inductance $$L$$:
$$ V_L = \omega L \cdot I $$
$$ 40 ~	ext{V} = \left(4 	imes 10^5 ~	ext{rad} ~	ext{s}^{-1}ight) \cdot L \cdot 0.5 ~	ext{A} $$
Rearrange to solve for $$L$$:
$$ L = \frac{40 ~	ext{V}}{(4 	imes 10^5 ~	ext{rad} ~	ext{s}^{-1}) \cdot 0.5 ~	ext{A}} $$
Simplify the expression:
$$ L = \frac{40}{2 	imes 10^5} $$
$$ L = 2 	imes 10^{-4} ~	ext{H} = 0.2 ~	ext{mH} $$
Therefore, the value of the inductance is $$0.2 ~	ext{mH}$$, which corresponds to option A.

Answer

$$0.2 \mathrm{~mH}$$

Answer

$$0.4 \mathrm{~mH}$$

Answer

$$0.8 \mathrm{~mH}$$

Answer

$$0.6 \mathrm{~mH}$$

AC fundamentals

A series L-C-R circuit containing a resistance of $120 ~\Omega has angular frequency 4 \times 10^5 \mathrm{~rad} \mathrm{~s}^{-1}. At resonance the voltage across resistance and inductor are 60 \mathrm{~V} and 40 \mathrm{~V}$ respectively, then the value of inductance will be

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