voltage across the inductor at resonance.

The given parameters are:

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:

$$ I = \frac{60 ~\text{V}}{120 ~\Omega} = 0.5 ~\text{A} $$

Now, use the value of current to solve for inductance $$L$$:

$$ V_L = \omega L \cdot I $$

$$ 40 ~\text{V} = \left(4 \times 10^5 ~\text{rad} ~\text{s}^{-1}\right) \cdot L \cdot 0.5 ~\text{A} $$

Rearrange to solve for $$L$$:

$$ L = \frac{40 ~\text{V}}{(4 \times 10^5 ~\text{rad} ~\text{s}^{-1}) \cdot 0.5 ~\text{A}} $$

Simplify the expression:

$$ L = \frac{40}{2 \times 10^5} $$

$$ L = 2 \times 10^{-4} ~\text{H} = 0.2 ~\text{mH} $$

Therefore, the value of the inductance is $$0.2 ~\text{mH}$$, which corresponds to option A.

Alternating Current

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