Home
Class 12
PHYSICS
An alternative voltage V=30 sin 50t+4...

An alternative voltage
V=30 sin 50t+40cos 50t is applied to a resitor of resistance 10`Omega`. Time rms value of current through resistor is

A

`5/sqrt2`

B

`10/sqrt2`

C

`7/sqrt2`

D

`7 A`

Text Solution

Verified by Experts

(a)`l=(V)/(R)=(30)/10 " sin"50t+(40)/(50)"cos"50t `
`or l=3 "sin "50t+4"cos" 50t `
` l=3 "sin "50t+4 "sin" (50t+(pi)/(2))`
`:. l_(rms)=l_(0)/sqrt2=5/sqrt2`
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • ALTERNATING CURRENT AND EM WAVE

    BITSAT GUIDE|Exercise BITSAT Archives|15 Videos

  • ATOMIC STRUCTURE

    BITSAT GUIDE|Exercise BITSAT Archives|4 Videos

Similar Questions

Explore conceptually related problems

An alternating voltage V=140 sin 50 t is applied to a resistor of resistance 10 Omega . This voltage produces triangleH heat in the resistor in time trianglet . To produce the same heat in the same time, rquired DC current is

Alternating voltage V = 400 sin ( 500 pi t ) is applied across a resistance of 0.2 kOmega . The r.m.s. value of current will be equal to

Knowledge Check

  • An A.C source of voltage V = 100 sin (100 pi t) is connected to a resistor of resistance 20 Omega . The rms value of current through resistor is

    A
    `10 A`
    B
    `10/(sqrt2) A`
    C
    `5/(sqrt2) A`
    D
    None of these

  • An alternating emf, e=300 sin100 pit volt, is applied to a pure resistance of 100 Omega . The rms current through the circuit is

    A
    2.12 A
    B
    0.212 A
    C
    20.12 A
    D
    0.0212 A

  • An alternating emf given by equation e=300sin(100pi)t V is applied to a resistance 100 Omega . The rms current through the circuit is (in amperes).

    A
    `3/sqrt(2)`
    B
    `9/sqrt(2)`
    C
    3
    D
    `6/sqrt(2)`

    • Similar Questions

    Explore conceptually related problems

    Alternating voltage V = 400 sin ( 500 pi t ) is applied across a resistance of 0.2 kOmega . The r.m.s. value of current will be equal to

    A current l=3+8 sin 100t is passing through a resistor of resistance 10 Omega .The effective value of current is

    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.

    An alternating voltage v(t) = 220 sin 100 pt volt is applied to a purely resistive load of 50Omega . The time taken for the current to rise from half of the peak value to the peak value is :

    In an AC circuit , an alternating voltage e = 200 sin 100t V is connected to a capacitor of capacity 1mu F . The rms value of the current in the circuit is

    Home
    Profile