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Why at resonance the voltage drop across inductance and the voltage drop across capacitance is greater than the source voltage?
Wiki User
∙ 12y ago
This isn't necessarily the case. It depends upon the value of resistance (which, at resonance, determines the current), and the values of the inductive- and capacitive-reactance.
At resonance, the impedance of the circuit is equal to its resistance. This is because the vector sum of resistance, inductive reactance, and capacitive reactance, is equal the the resistance. This happens because, at resonance, the inductive- and capacitive-reactance are equal but opposite. Although they still actually exist, individually.
If the resistance is low in comparison to the inductive and capacitive reactance, then the large current will cause a large voltage drop across the inductive reactance and a large voltage drop across the capacitive reactance. Because these two voltage drops are equal, but act the opposite sense to each other, the net reactive voltage drop is zero.
So, at (series) resonance:
a. the circuit's impedance is its resistance (Z = R)
b. the current is maximum
c. the voltage drop across the resistive component is equal to the supply voltage
d. the voltage drop across the inductive-reactance component is the product of the supply current and the inductive reactance
e. the voltage drop across the capacitive-reactance component is the product of the supply current and the capacitive reactance
f. the voltage drop across both inductive- and capacitive-reactance is zero.
Wiki User
∙ 12y ago
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