Question

A coil of inductanceand $\mathbf{88}\mathbf{}\mathbf{\text{mH}}\mathbf{}$unknown resistance and a $\mathbf{0}\mathbf{.}\mathbf{94}\mathbf{\text{μF}}$capacitor are connected in series with an alternating emf of frequency $\mathbf{930}\mathbf{}\mathbf{}\mathbf{\text{Hz}}$. If the phase constant between the applied voltage and the current is $\mathbf{75}\mathbf{°}$, what is the resistance of the coil?

Short answer

The resistance of the coil is $89\text{Ω}$.

Step-by-step solution

The given data

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

Understanding the concept of phase angle

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 $\varphi$in the sinusoidally driven $RLC$circuit

$\tan \varphi =\frac{X_L-X_c}{R}$ ...(i)

The inductive reactance of an inductor,

$X_L={\omega }_{d}L$ ...(ii)

The capacitive reactance of a capacitor,

$X_C=\frac{1}{{\omega }_{d}C}$ ...(iii)

The angular frequency of an oscillation,

${\omega }_d=2\pi f_d$ ...(iv)

Calculation of the resistance of the coil

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

$tan\varphi =\frac{{\omega }_{d}L-\frac{1}{{\omega }_{d}C}}{R}$

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

role="math" localid="1662756970044" $$\begin{gathered} R=\frac{1}{tan\varphi }\left({\omega }_{d}L-\frac{1}{{\omega }_{d}C}\right) \\ =\frac{1}{tan\varphi }\left(2\pi f_{d}L-\frac{1}{2\pi f_{d}C}\right) \\ =\frac{1}{tan75°}\left[2\pi \left(930\text{Hz}\right)\left(8.8\times {10}^{-2}\text{H}\right)-\frac{1}{2\pi \left(930\text{Hz}\right)\left(0.94\times {10}^{-6}\text{F}\right)}\right] \\ =89\text{Ω} \end{gathered}$$

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