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Chapter 4: 37P. (page 1046)

A coil has a resistance of 48.0 Ω. At a frequency of 80.0 Hz the voltage across the coil leads the current in it by 52.3°. Determine the inductance of the coil.

Short Answer

Expert verified

The value of inductance L in the circuit is 0.124H

Step by step solution

01

Step-1: Formulas used   

\(\phi \) is the angle between the voltage and current phasors

\(\tan \phi = \frac{{({X_L} - {X_C})}}{R}\),where \({X_C}\) is the capacitive reactance, \({X_L}\)is the inductive reactance and R is the resistance.

The circuit does not have a capacitor so \({X_C}\)= 0 so,

\({X_L} = R\tan \phi \)

To calculate L use,

\(L = \frac{{{X_L}}}{{2\pi f}}\)

02

Step-2: Calculations for Inductance

Plug the values of R and phase difference to get \({X_L}\)

\(\begin{aligned}{X_L} = (48\Omega )\tan (52.3^\circ )\\ = 62.1\Omega \end{aligned}\)

Now, plug the values of \({X_L}\)and f to get L

\(\begin{aligned}L = \frac{{62.1\Omega }}{{2\pi (80Hz)}}\\ = 0.124H\end{aligned}\)

Therefore the value of inductance of coil L = 0.124H

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