91Ó°ÊÓ

1. Inductance of the coil, $L=88\text{mH}$
2. Capacitance of the capacitor,$C=0.94\text{μ¹ó}$
3. Frequency of the alternating emf,$f_d=930\text{Hz}$
4. Phase constant between the applied voltage and current,$\mathrm{ϕ}=75°$

Phase angle is the angle between the voltage and current phasor. It represents the phase difference between voltage and current in the circuit. We use the relation of phase constant with resistance $\left(R\right)$, Inductance $\left(L\right)$and capacitance $\left(C\right)$. Substituting the value of capacitive reactance, inductive reactance, and phase constant, we can find the resistance of the coil.

An equation for the phase constant $\mathrm{Ï•}$in the sinusoidally driven $RLC$circuit

$\tan \mathrm{Ï•}=\frac{X_L-X_c}{R}$ ...(i)

The inductive reactance of an inductor,

$X_L={\mathrm{Ó\neg }}_{d}L$ ...(ii)

The capacitive reactance of a capacitor,

$X_C=\frac{1}{{\mathrm{Ó\neg }}_{d}C}$ ...(iii)

The angular frequency of an oscillation,

${\mathrm{Ó\neg }}_d=2\mathrm{Ï€}{f}_d$ ...(iv)

Substituting the values of equations (ii) and (iii) in equation (i), and rearranging it we can get the resistance of the coil as follows:

$tan\mathrm{Ï•}=\frac{{\mathrm{Ó\neg }}_{d}L-\frac{1}{{\mathrm{Ó\neg }}_{d}C}}{R}$

For the given values, the above equation can also be written as-

role="math" localid="1662756970044" $$\begin{gathered} R=\frac{1}{tan\mathrm{Ï•}}\left({\mathrm{Ó\neg }}_{d}L-\frac{1}{{\mathrm{Ó\neg }}_{d}C}\right) \\ =\frac{1}{tan\mathrm{Ï•}}\left(2\mathrm{Ï€}{f}_{d}L-\frac{1}{2\mathrm{Ï€}{f}_{d}C}\right) \\ =\frac{1}{tan75°}\left[2\mathrm{Ï€}\left(930\text{Hz}\right)\left(8.8×{10}^{-2}\text{H}\right)-\frac{1}{2\mathrm{Ï€}\left(930\text{Hz}\right)\left(0.94×{10}^{-6}\text{F}\right)}\right] \\ =89\text{Ω} \end{gathered}$$

Hence, the value of the resistance is $89\text{Ω}$.