Alternating Current Test - 39

When you join your new assignment, on the very first day your boss gives you an inductor and asks you to find its inductance. You are provided an AC voltmeter of high impedance, a resistor, a capacitor and an AC source of variable frequency. You recall from your physics class that if R, L and C are connected in series current is maximum at resonance. Since you are not provided an ammeter yon drop this idea. To measure inductance, however, on the advice of an experienced colleague you connect L, R and C in series with the AC source of variable frequency.

Under certain conditions voltages as small as 10 V can be dangerous, and should not be regarded with anything but respect and caution.
On the positive side, rapidly alternating currents can have beneficial effects. Alternating currents with frequencies of the order of $$10^6 Hz$$ do not interfere appreciably with nerve processes, and can be used for therapeutic heating for arthritic conditions, sinusitis, and a variety of other disorders. If one electrode is made very small, the resulting concentrated heating can be used for local destruction of tissue, such as tumors, or even for cutting tissue in certain surgical procedures.
Study of particular nerve impulses is also an important diagnostic tool in medicine. The most familiar examples are electrocardiography (ECG) and electro encephalography (EEG). Electrocardiograms, obtained by attaching electrodes to the chest and back and recording the regularly varying potential differences, are used to study heart function. Similarly, electrodes attached to the scalp permit study of potentials in the brain and the resulting patterns can be helpful in diagnosing epilepsy, brain tumors, and other disorders.

The current in an $$L-C-R$$ series circuit fed by an alternating voltage source $$V =V_0\sin \omega t$$ is given by
$$\displaystyle i = \dfrac{V_0\sin (\omega t - \phi)}{\sqrt{R^2 + \left(L\omega - \displaystyle \dfrac{1}{C\omega}\right)^2}}.....................(1)$$

where $$\displaystyle \tan\phi = \frac{\omega L- \dfrac{1}{\omega C}}{R} .....................................(2)$$

$$\phi$$ is the phase difference between the current and the voltage. The average power delivered by the source to the circuit is given by the equation 

$$\displaystyle P= \frac{1}{2} \frac{V_0^2 cos \phi}{|Z|}$$.........(3)

Where |Z| is the impedance of the circuit given by the denominator of equation (1).

In a series LCR circuit with an ideal ac source of peak voltage $${E}_{0}=50V$$, frequency $$v=\cfrac{50}{\pi}$$ $$Hz$$ and $$R=300\Omega$$. The average electric field energy stored in the capacitor and average magnetic energy stored in the coil are $$25mJ$$ and $$5mJ$$ respectively. The value of RMS current in the circuit is $$0.1A$$. Then find: