I would like to know what would happen if a capacitive load is connected to the secondary of a transformer.
I simulated the circuit on Proteus. The inductance of the secondary coil was 0.1 H
and the frequency of the supply is 50 Hz. The secondary coil produces 15V peak to peak.
I connected a capacitor across the terminals of the secondary coil.
When the capacitive reactance of the load is equal to the inductive reactance of the secondary coil (Resonance at 50 Hz), I found that the voltage is increased to about 50 V peak to peak!! (The value of the capacitor was 100uF or 101uF).
First, I thought that the mistake was in the simulator so I connected a small resistance in series with the capacitor. I tried from 1 ohm to 50 ohm. The voltage was dropped but it was sill higher than 15V peak to peak.
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Why resonance increases the voltage?
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If I need to increase the voltage, Why should the resonance be at 50 Hz which is the frequency of the supply? Why the voltage does not increase at a frequency of say 2 KHz?
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If I made this experiment practically, Will the transformer wear out?
Thank you very much,
simulate this circuit – Schematic created using CircuitLab
Best Answer
The resonant frequency of an LC circuit is given by
$$ f = \frac {1}{2 \pi \sqrt {LC}} $$
and plugging in the values you have chosen gives
$$ f = \frac {1}{2 \pi \sqrt {0.1 \cdot 100 \mu}} = \frac {1}{2 \pi \cdot 0.00316 } = 50.3~Hz$$
So your resonant frequency is almost exactly on the mains frequency of 50 Hz.
It's like pushing a swing or pendulum. If you do it at the natural frequency of oscillation and at the right point in the cycle then every push will boost a little bit more. In the case of the pendulum or swing air resistance will eventually limit the amplitude of the oscillation. In the LC circuit the internal resistance will cause losses which increase with increasing amplitude until an equilibrium is reached.
See the formula.
Transformers lifespan is mostly determined by temperature which is usually a result of too-much current or insulation breakdown which is a result of too high a voltage being applied. You would need to know the winding insulation breakdown voltage to figure that out. At 50 V you are unlikely to have a problem.