Answer
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Hint: In order to solve this question, we need to use the formula of power dissipation in an LCR circuit which is given as, Power =${V}_{rms}.{I}_{rms}.cos\phi$. Then, we need to simplify this equation and we will reach the conclusion.
Formulae used:
Power dissipated in an LCR circuit, Power$={V}_{rms}.{I}_{rms}.cos\phi$, where ${V}_{rms}$ is the root-mean-square voltage, ${I}_{rms}$ is the root-mean-square current and $cos\phi$ is the power factor.
Impedance, $Z=\sqrt{R^2 + ({X}_{L}-{X}-{C})^2}$, where R is the DC resistance, ${X}_{L}$ is the inductive reactance and ${X}_{C}$ is the capacitive reactance.
Complete step-by-step answer:
Let us consider an RLC circuit with resistance R, capacitance C and inductance I. We know that, in an RLC circuit, the power dissipated is given by Power$={V}_{rms}.{I}_{rms}.cos\phi$, where ${V}_{rms}$ is the root-mean-square voltage, ${I}_{rms}$ is the root-mean-square current and $cos\phi$ is the power factor.
And impedance is given by $Z=\sqrt{R^2 + ({X}_{L}-{X}_{C})^2}$, where R is the DC resistance, ${X}_{L}$ is the inductive reactance and ${X}_{C}$ is the capacitive reactance
Power factor, $cos\phi=\dfrac{R}{Z}$
So, we can now write, Power$={V}_{rms}.{I}_{rms}.\dfrac{R}{Z}={I}_{rms}Z.{I}_{rms}.\dfrac{R}{Z}={I}_{rms}^2 .R$
Thus, we can see that power dissipation only depends on the root-mean-square current and the resistance of the circuit.
Hence, option a is the correct answer.
Additional Information:
Impedance is the total effective resistance of an electric circuit to the alternating current which arises from the combined presence of resistance and reactance.
Inductive reactance and capacitive reactance is given by ${X}_{L}=\omega L$ and ${X}_{C}=\dfrac{1}{\omega C}$ respectively, where $\omega$ is the angular velocity of the AC circuit.
Note: Sometimes, it may be asked in the question about the part of the circuit from where the energy is going to be dissipated. In that case, we must know that power dissipation and energy dissipation both occur in the resistor only as we know that in a series resonating LCR circuit, the energy keeps oscillating from the inductor to capacitor.
Formulae used:
Power dissipated in an LCR circuit, Power$={V}_{rms}.{I}_{rms}.cos\phi$, where ${V}_{rms}$ is the root-mean-square voltage, ${I}_{rms}$ is the root-mean-square current and $cos\phi$ is the power factor.
Impedance, $Z=\sqrt{R^2 + ({X}_{L}-{X}-{C})^2}$, where R is the DC resistance, ${X}_{L}$ is the inductive reactance and ${X}_{C}$ is the capacitive reactance.
Complete step-by-step answer:
Let us consider an RLC circuit with resistance R, capacitance C and inductance I. We know that, in an RLC circuit, the power dissipated is given by Power$={V}_{rms}.{I}_{rms}.cos\phi$, where ${V}_{rms}$ is the root-mean-square voltage, ${I}_{rms}$ is the root-mean-square current and $cos\phi$ is the power factor.
And impedance is given by $Z=\sqrt{R^2 + ({X}_{L}-{X}_{C})^2}$, where R is the DC resistance, ${X}_{L}$ is the inductive reactance and ${X}_{C}$ is the capacitive reactance
Power factor, $cos\phi=\dfrac{R}{Z}$
So, we can now write, Power$={V}_{rms}.{I}_{rms}.\dfrac{R}{Z}={I}_{rms}Z.{I}_{rms}.\dfrac{R}{Z}={I}_{rms}^2 .R$
Thus, we can see that power dissipation only depends on the root-mean-square current and the resistance of the circuit.
Hence, option a is the correct answer.
Additional Information:
Impedance is the total effective resistance of an electric circuit to the alternating current which arises from the combined presence of resistance and reactance.
Inductive reactance and capacitive reactance is given by ${X}_{L}=\omega L$ and ${X}_{C}=\dfrac{1}{\omega C}$ respectively, where $\omega$ is the angular velocity of the AC circuit.
Note: Sometimes, it may be asked in the question about the part of the circuit from where the energy is going to be dissipated. In that case, we must know that power dissipation and energy dissipation both occur in the resistor only as we know that in a series resonating LCR circuit, the energy keeps oscillating from the inductor to capacitor.
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