Assertion: No power loss associated with pure capacitor in ac circuit.
Reason: No current is flowing in this circuit.
A. If both assertion and reason is true but the reason is the correct explanation of assertion.
B. If both assertion and reason is true but the reason is not the correct explanation of assertion.
C. If assertion is true but reason is false.
D. If both assertion and reason are false.
E. If reason is true but assertion is false.
Answer
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Hint: We will observe a circuit that has only two elements: an Ac Voltage source and a pure capacitor with capacitance C. When a capacitor is connected to an Ac source, capacitors continue to charge and discharge simultaneously with respect to the change in voltage source. A pure capacitor means it has no real resistance but has an impedance Z. And we will try to find current flowing in the circuit and its relation to Voltage.
Complete answer:
We are assuming that voltage of the source is \[V={{V}_{0}}\sin \omega t\], where \[{{V}_{0}}\]is amplitude of the voltage and \[\omega \] is angular frequency. Now for a pure capacitor with capacitance C, have impedance Z=j/\[\omega \]C, where \[{{j}^{2}}=-1\], which means it is purely imaginary. We can represent it as amplitude with a phase. Now to calculate the current we will use ohm’s law and divide the voltage with impedance. \[I={{V}_{0}}\sin \omega t/Z={{V}_{0}}\omega C\sin (\omega t+\dfrac{\Pi }{2})\].
As you can see, the circuit will have a flowing current. But it will not dissipate any power because the circuit does not have any dissipative element like a pure resistor or any element which has a real part of resistance through which power will be dissipated.
The Assertion is true but reason is false.
The correct answer will be option C.
Note:
The current of the system will always be lagging to its voltage because of the phase of capacitor impedance. The phase difference will be\[\dfrac{\Pi }{2}\]. In most of the circuit with real resistors, the power will dissipate by heat energy. In some cases, it may be as radiation energy.
Complete answer:
We are assuming that voltage of the source is \[V={{V}_{0}}\sin \omega t\], where \[{{V}_{0}}\]is amplitude of the voltage and \[\omega \] is angular frequency. Now for a pure capacitor with capacitance C, have impedance Z=j/\[\omega \]C, where \[{{j}^{2}}=-1\], which means it is purely imaginary. We can represent it as amplitude with a phase. Now to calculate the current we will use ohm’s law and divide the voltage with impedance. \[I={{V}_{0}}\sin \omega t/Z={{V}_{0}}\omega C\sin (\omega t+\dfrac{\Pi }{2})\].
As you can see, the circuit will have a flowing current. But it will not dissipate any power because the circuit does not have any dissipative element like a pure resistor or any element which has a real part of resistance through which power will be dissipated.
The Assertion is true but reason is false.
The correct answer will be option C.
Note:
The current of the system will always be lagging to its voltage because of the phase of capacitor impedance. The phase difference will be\[\dfrac{\Pi }{2}\]. In most of the circuit with real resistors, the power will dissipate by heat energy. In some cases, it may be as radiation energy.
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