An inductor of self inductance 1 H connected in series with a resistor of 100 and an ac supply of 100 volt, 50 Hz. Maximum current flowing in the circuit is ________ A. [2025]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
An inductor of self inductance 1 H connected in series with a resistor of 100 $π Ω$ and an ac supply of 100 $π$ volt, 50 Hz. Maximum current flowing in the circuit is ________ A. [2025]

Ans.
(1)

Impedance of circuit

$Z = \sqrt{R^{2} + {(X_{L})}^{2}} = \sqrt{R^{2} + {(ω L)}^{2}}$

$= \sqrt{{(100 π)}^{2} + {(2 π \times 50 \times 1)}^{2}}$

$= \sqrt{{(100 π)}^{2} + {(100 π)}^{2}}$

$= \sqrt{2} \times 100 π Ω$

$I_{rms} = \frac{V}{Z} = \frac{100 π}{\sqrt{2} \times 100 π} = \frac{1}{\sqrt{2}}$

$I_{\max} = \sqrt{2} I_{rms} = \sqrt{2} \times \frac{1}{\sqrt{2}} = 1 Ampere$

Want Us to Email you About Special Offers & Updates?