
In which of the following circuits the maximum power dissipation is observed?
$\left( {\text{A}} \right)$ Pure capacitive circuit
$\left( {\text{B}} \right)$ Pure inductive circuit
$\left( {\text{C}} \right)$ Pure resistive circuit
$\left( {\text{D}} \right)$ None of these
Answer
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Hint: Power dissipation: It is defined as the rate at which energy is lost due to resistive elements in the circuit.
Power: It is the amount of work done per unit time.
If we know the voltage drop across a component and the current through it, we can calculate the power dissipation by using Ohm’s law.
Formula used:
By using Ohm’s law, ${\text{V = IR}}$, here\[\;V = \] Potential difference across the circuit, \[I = \] current flowing in the circuit,
\[R = \] Resistance of the circuit.
Power dissipation,${\text{P = }}{{\text{V}}_{{\text{rms}}}}{{\text{I}}_{{\text{rms}}}}{\text{cos}}\phi $, here \[{V_{rms}} = \] RMS voltage difference across the circuit,
\[{I_{rms}} = \] RMS current flow in the circuit, $\phi = $ phase difference between the current and the voltage
Complete step by step answer:
In a purely resistive and capacitive circuit, the phase difference is \[{90^ \circ }\]-,
By substituting the values of $\phi $ in the equation we get,
${\text{P = }}{{\text{V}}_{{\text{rms}}}}{{\text{I}}_{{\text{rms}}}}{\text{cos}}\phi $
=${{\text{V}}_{{\text{rms}}}}{{\text{I}}_{{\text{rms}}}}{\text{cos9}}{{\text{0}}^0} = 0$, as we know that the value of ${\text{cos9}}{{\text{0}}^0} = 0$
But in a purely resistive circuit, the voltage and current are in the same phase so phase difference is zero. Therefore, it will have maximum power dissipation when, ${\text{cos}}\phi {\text{ = 1}}$, that means $\phi = {0^0}$.
Hence, the correct option is $\left( {\text{C}} \right)$.
Note:
RMS voltage: It is the square root of the mean value of the square function of the instantaneous values of potential difference or the voltage.
RMS current: It is the square root of the mean value of the square function of the instantaneous values of current.
The pure capacitive circuits only contain a capacitor in the circuit.
The pure inductive circuits only contain an inductor in the circuit.
The pure resistive circuits only contain a resistor in the circuit.
Power: It is the amount of work done per unit time.
If we know the voltage drop across a component and the current through it, we can calculate the power dissipation by using Ohm’s law.
Formula used:
By using Ohm’s law, ${\text{V = IR}}$, here\[\;V = \] Potential difference across the circuit, \[I = \] current flowing in the circuit,
\[R = \] Resistance of the circuit.
Power dissipation,${\text{P = }}{{\text{V}}_{{\text{rms}}}}{{\text{I}}_{{\text{rms}}}}{\text{cos}}\phi $, here \[{V_{rms}} = \] RMS voltage difference across the circuit,
\[{I_{rms}} = \] RMS current flow in the circuit, $\phi = $ phase difference between the current and the voltage
Complete step by step answer:
In a purely resistive and capacitive circuit, the phase difference is \[{90^ \circ }\]-,
By substituting the values of $\phi $ in the equation we get,
${\text{P = }}{{\text{V}}_{{\text{rms}}}}{{\text{I}}_{{\text{rms}}}}{\text{cos}}\phi $
=${{\text{V}}_{{\text{rms}}}}{{\text{I}}_{{\text{rms}}}}{\text{cos9}}{{\text{0}}^0} = 0$, as we know that the value of ${\text{cos9}}{{\text{0}}^0} = 0$
But in a purely resistive circuit, the voltage and current are in the same phase so phase difference is zero. Therefore, it will have maximum power dissipation when, ${\text{cos}}\phi {\text{ = 1}}$, that means $\phi = {0^0}$.
Hence, the correct option is $\left( {\text{C}} \right)$.
Note:
RMS voltage: It is the square root of the mean value of the square function of the instantaneous values of potential difference or the voltage.
RMS current: It is the square root of the mean value of the square function of the instantaneous values of current.
The pure capacitive circuits only contain a capacitor in the circuit.
The pure inductive circuits only contain an inductor in the circuit.
The pure resistive circuits only contain a resistor in the circuit.
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