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

Question

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

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

Monjur Alahi

5 views

2025-01-04 04:42

1 Answers

Related Questions

A current in a circuit is due to a potential difference of 30 V applied to a resistor of resistance `300 Omega`. What resistance would permit the same

2 Answers 4 Views

An L-C_R series circuit is joined to a source of alternating e.m.f. If `R=9 Omega, X_(L)=28 Omega, X_(C)=16 Omega`, then the impedance of the circuit

2 Answers 4 Views

An alternating e.m.f. `E=10sqrt(2) sin omega` volt is applied to a circuit containing a pure inductance L and a resistance and voltage acros the resis

2 Answers 5 Views

An inductance of 100 mH and a resistance of 100 `Omega` are connected in series and an alternating emf of peak value 100 V, 50 Hz is applied across th

2 Answers 5 Views

An alternating emf, e=200 sin `omegat` is applied to a lamp, whose filament has a resistance of `1000Omega`. The rms valve of current is

2 Answers 8 Views

When an alternating emf, e=300 sin `100pit` volt is applied across a bulb, the peak value of current is found to be 2 A. the average power is

2 Answers 4 Views

An alternating emf of peak value 350 V is applied across an a.c. ammeter of resistance `100 Omega`. What is the reading of the ammeter?

2 Answers 5 Views

An alternating emf of 200 V, 50 Hz is applied a circuit containing a resistance of 100 `Omega` and inductance of 0.1 H in series. The powerr dissipate

2 Answers 4 Views

In LCR series circuit , an alternating emf e and current `i` are given by the equations `e = 100sin (100 t) ` volt . `i=100 sin (100 t + (pi)/(3))` mA

2 Answers 5 Views

An alternating supply of 220 V is applied across a circuit with resistance `22 Omega` and impedance `44 Omega`. The power dissipated in the circuit is

2 Answers 5 Views

Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - A
Comparing given equation with
`e=e_(0)sin omegat`, we have, `e_(0)=300V`
`therefore I_(rms)=(I_(0))/(sqrt(2))`
`=(e_(0))/(Rsqrt(2))=(300)/(100xx1.414)=2.12A`

5 views Answered 2023-02-05 15:29