Let an inductor of inductance L, a resistance of resistance R and a capacitor of capacitance C be connected in series across an ac source of frequency f.
Let the current vary as
Where I is the current amplitude and
We know that the potential difference across the resistor is given by:
where VRm is the maximum potential across the capacitor and
We know that the potential difference across the capacitor is given by:
where VCm is the maximum potential across the capacitor
We know that the potential difference across the inductor is given by:
where VLm is the maximum potential across the capacitor and
Drawing the phasor diagram,
We can find the total potential difference across the three components by adding them. We do scalar addition for these potentials. This is same as adding the horizontal components of the phasors. Hence, we can do the vector addition of the phasors to get a new phasor V . The horizontal component of this phasor will give us the total potential difference across the three components.
Now, using the Pythagoras theorem, we have
Substituting , and , we have
Impedance, Z, is defined as
Note that the phase angle is the term inside the sine or the cosine.
The phase angle for the capacitor is
and the phase angle for the inductor is
Hence, the phase difference is π radians or 180° .
In a dc circuit, an inductor acts as a short circuit. Hence, it has negligible resistance. However, in an ac circuit, an inductor provides significant resistance known as inductive reactance. Thus, the current reduces.
Let us try to find the self-inductance.
Note that the inductor has some resistance. Hence, it can be though of as an ideal inductor in series with a resistor.
In the first circuit,
The resistance is given by:
(Note that the inductor acts as a short circuit)
(a) Such a device is known as a step-down transformer.
It is based on the principle of induction. If the current is changing through one coil, current is induced in a nearby coil.
The four sources of energy loss are flux leakage, resistance of the windings, eddy currents and hysteresis loss.
The town has power demand of 12000kW at 220V.
Resistance per unit length
Length of each wire
The transformer steps down 4000V () to 220V( ).
If there is no power loss in the transformer, we can calculate the rms current in the lines by:
The power loss in the lines is
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