A 110 V, 60 Hz AC line is connected to a 6.55 mH coil. Find the current through the coil.
Xl = 6.28*60*(6.55*10^-3h) = 2.47 Ohms.
= Reactance of coil.
I=V / Xl = 110 / 2.47 = 44.5Amps., rms.
The impedance is
Z = 2 pi f L
where f = 60 Hz and L = 6.55*10^-3 H
= 2.47 Ohms
The peak voltage is
Vmax = Vsqrt2 = 155.6 V
Current will lag voltage by 90 degrees,
and the peak current will be
Vmax/Z = 63.0 Amperes
The resistance of the coil has been neglected.
My Imax result is equivalent to Henry's Irms result. The ratio of the two is sqrt2
Thank u! I was halfway there.
To find the current through the coil, we need to use Ohm's law and the equation for the impedance of an inductor in an AC circuit.
First, let's find the impedance of the coil, which represents the opposition offered by the coil to the flow of current:
Z = 2πfL,
Where:
Z = impedance of the coil,
π ≈ 3.14159,
f = frequency of the AC line (60 Hz),
L = inductance of the coil (6.55 mH or 6.55 × 10^(-3) H).
Substituting the known values into the formula, we get:
Z = 2π × 60 × 6.55 × 10^(-3).
Calculate the value of Z using a calculator:
Z ≈ 2π × 60 × 6.55 × 10^(-3) ≈ 0.778 Ω.
Now, to find the current through the coil, we can apply Ohm's law:
I = V/Z,
Where:
I = current through the coil,
V = voltage of the AC line (110 V),
Z = impedance of the coil (0.778 Ω).
Substituting the values into the formula, we get:
I ≈ 110/0.778.
Calculate the value of I using a calculator:
I ≈ 110/0.778 ≈ 141.36 A.
Therefore, the current through the coil is approximately 141.36 A.