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With proper phase diagrams, explain behaviour of a pure capacitor ina AC circuit.
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written 3.3 years ago by
teamques10
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(1) $Z=R+(X_L-X_C)j =-jX_C = X_C\angle -90$
(2) $I=\dfrac{V}{Z}=\dfrac{V\angle 0}{X_C\angle -90}=\left ( \dfrac{V}{X_C} \right )\angle 0 -(-90)\\[2ex] I=I_C\angle 90$
(3) In a pure capacitive circuit, current leads the voltage by $90^\circ$
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