Home
Class 12
PHYSICS
An a.c. source of voltage V = 100 sin 10...

An a.c. source of voltage `V = 100 sin 100 pi t`. Is connected to a resistor of `(25)/(sqrt2) ohm`. What is the r.m.s value of current (in ampere) through the resistor ?

Text Solution

Verified by Experts

The correct Answer is:
4
Promotional Banner

Topper's Solved these Questions

  • ELECTROMAGNETIC OSCILLATIONS AND ALTERNATING CURRENT

    RESNICK AND HALLIDAY|Exercise Practice Questions (Linked Comprehension)|11 Videos
  • ELECTROMAGNETIC INDUCTION

    RESNICK AND HALLIDAY|Exercise PRACTICE QUESTIONS (Integer Type)|14 Videos
  • ELECTROMAGNETIC WAVES

    RESNICK AND HALLIDAY|Exercise PRACTICE QUESTIONS (Integer Type)|3 Videos

Similar Questions

Explore conceptually related problems

An Ac source of volatage V=100 sin 100pit is connected to a resistor of ressistance 20 Omega The rsm value of current through resistor is

An Ac source of volatage V=100 sin 100pit is connected to a resistor of ressistance 20 Omega The rsm value of current through resistor is , power factor is

An Ac source of volatage V=100 sin 100pit is connected to a resistor of ressistance 20 Omega The rsm value of current through resistor is total charge transferred through resistor in long time is

An Ac source of volatage V=100 sin 100pit is connected to a resistor of ressistance 20 Omega The rsm value of current through resistor is average value of current for half cycle is

An Ac source of volatage V=100 sin 100pit is connected to a resistor of ressistance 20 Omega The rsm value of current through resistor is ,total charge transferred in 1/100 second is

An Ac source of volatage V=100 sin 100pit is connected to a resistor of ressistance 20 Omega The rsm value of current through resistor is the averge value for half cycle is

An Ac source of volatage V=100 sin 100pit is connected to a resistor of ressistance 20 Omega The rsm value of current through resistor is ,total heat generated in one cycle is

An alternating voltage given by V=70 sin 100 pi t is connected across a pure resistor of 25 Omega. Find (i) the frequency of the source. (ii) the rms current through the resistor.

A alternating voltage given by V = 140 sin 314 t is connected across a pure resistor of 50 ohm. Find the rms current through the resistor.