Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

an inductor of 3h is connected to abattery of emf 6v through a resistance of 100calculate the time constant what will be the maxmimum value of current in the circuit.

Answers (1)

 

Give that L=3H,E=6V,R=100Ω
Time constant \tau _{L} = \frac{L}{R}= \frac{3}{100}= 0.03 \ sec
Maximum Current I_0= \frac{E}{R} = \frac{6}{100} = 0.06 \ amp

Posted by

Satyajeet Kumar

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2026: IIT Roorkee’s ECE emerges as top choice after CSE; closing rank rises
anam-khan

JEE Main 2026 application correction begins on jeemain.nta.nic.in; edit details by December 2
anam-khan

JEE Mains 2026 LIVE: NTA to conduct session 1 from January 21 to 30; exam pattern, syllabus

Latest Question