A 0.21 H inductor and a 12 ohm resistanc...

A `0.21 H` inductor and a `12 ohm` resistance are connected in series to a `220 V`. `50 Hz` ac source. Calculate the current in the circuit and the phase angle between the current and the source voltage.

Text Solution

Verified by Experts

`X_(L) = omega L = 2 pi fL = 2 pi xx 50 xx 0.21 = 21 pi Omega`
`Z = sqrt(R^(2) + X_(L)^(2)) = sqrt(12^(2) + (21 pi)^(2)) = sqrt(144 + 4348)`
`Z = sqrt(4492) ~= 67.01 Omega , I + (V)/(Z) = (220)/(67.02) = 3.28 A`
`phi = tan^(-1) ((X_(L))/(R )) = tan^(-1) ((21 pi)/(12))`
The current will lag applied voltage by an angle `tan^(-1) ((21 pi)/(12))`

Topper's Solved these Questions

ALTERNATING CURRENT
NARAYNA|Exercise C.U.Q|66 Videos
ALTERNATING CURRENT
NARAYNA|Exercise Assertion & Reason|14 Videos
ATOMIC PHYSICS
NARAYNA|Exercise LEVEL-II (H.W)|14 Videos

A 0.21-H inductor and a 88-Omega resistor are connected in series to a 220-V, 50-Hz AC source. The current in the circuit and the phase angle between the current and the source voltage are respectively. (Use pi= 22//7)

`2A, tan^(-1)3//4`

`14.4A, tan^(-1)7//8`

`14.4A, tan^(-1)8//7`

`3.28A, tan^(-1)2//11`

A 0.21 H inductor and a 12 Omega resistance are connected in series to a 220 V, 50 Hz ac source. The current in the circuit is

`(220)/(sqrt(4400))A`

`(22)/(3sqrt(5)) A`

`(220)/(sqrt(4500))A`

`(22)/(5sqrt(3)) A`

A 0.07 H inductor and a 12 Omega resistor are connected in series to a 220 V, 50 Hz ac source. The approximate current in the circuit and the phase angle between current and source voltage are respectively. [Take pi as (22)/(7) ]

`8.8A and tan^(-1)((11)/(6))`

`88A and tan^(-1)((11)/(6))`

`0.88A and tan^(-1)((11)/(6))`

`8.8 A and tan^(-1)((6)/(11))`