A bulb of resistance 10 Ω, connected to an inductor of inductance L, is in series with a source marked 100 V, 50 Hz. If the phase angle between the voltage and current is π/4 radian. The value of L will be:

Concept:

LR Circuit:

• An LR Circuit is an electrical circuit consisting of an inductor (L), resistor (R) it can be connected either parallel or series.
• Let the applied voltage be V = V₀Sinωt, then the instantaneous current leads the voltage by an angle ϕ then
• The instantaneous voltage and current  is given by

\( \Rightarrow e = {e_0}\sin \omega t\)

\( \Rightarrow I = {I_0}\sin \left( {\omega t - \phi } \right)\)

The phase angle is given by:-

\(\phi = {\tan ^{ - 1}}\left( {\frac{{{X_L}}}{R}} \right)\)

Where X_L = Inductive reactance, R= Resistance

Calculation:

Calculate the value of L

\(\tan \phi = \frac{{{X_L}}}{R} = \frac{{\omega L}}{R} = \frac{{2\pi fL}}{R}\)

\(L = \frac{{R\tan \phi }}{{2\pi f}}\)

Now, \(\phi = 45^\circ \;,\tan \phi = 1\)

f = 50 Hz, R = 10 Ω

\(\therefore L = \frac{{10\; \times \;1\; \times \;7}}{{2\; \times\; 22\; \times\; 50}}\;H\)

= 0.032 H