1
Question
An inductor of reactance \(1~\Omega\) and a resistor of \(2~\Omega\)are connected in series to the terminals of a \(6~π(πππ )\) AC source.
The power dissipated in the circuit is:
The power dissipated in the circuit is:
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Solution
Formula used:
\(π=π_{πππ } πΌ_{ππ}~ πππ β‘\phi=π_{πππ } πΌ_{πππ }\dfrac{ π
}{π}\)
Given,
Inductive reactance, \(π_{l}=1~\Omega\)
Capacitive reactance, \(π_{πΆ}=0\)
Resistance, \(π
=2~ \Omega\)
Rms voltage, \(π_{πππ }=6~ π\)
As we know, Impedance of the circuit,
\(π=\sqrt{π_{πΏ}^{2}+π
^{2}}=\sqrt{(1)^{2}+(2)^{2}}\)
\(π =\sqrt{5}~ \Omega\)
The average power dissipated by resistor (power is never dissipated by inductor or capacitor) of resistance π
in any ac circuit with πππ current \(πΌ_{πππ }\) is given by,
\(πΌ_{πππ }=\dfrac{π_{πππ }}{π}=\dfrac{6}{\sqrt{5}} ~π΄\)
Power dissipated,
\(π=π_{πππ } πΌ_{ππ}~ πππ β‘\phi=π_{πππ } πΌ_{πππ }\dfrac{ π
}{π}\)
\(π=6\times\dfrac{6}{\sqrt{5}}\times\dfrac{2}{\sqrt{5}}=\dfrac{72}{5}=14.4 ~π\)
Final Answer: Option (c)
\(π=π_{πππ } πΌ_{ππ}~ πππ β‘\phi=π_{πππ } πΌ_{πππ }\dfrac{ π }{π}\)
Given,
Inductive reactance, \(π_{l}=1~\Omega\)
Capacitive reactance, \(π_{πΆ}=0\)
Resistance, \(π =2~ \Omega\)
Rms voltage, \(π_{πππ }=6~ π\)
As we know, Impedance of the circuit,
\(π=\sqrt{π_{πΏ}^{2}+π ^{2}}=\sqrt{(1)^{2}+(2)^{2}}\)
\(π =\sqrt{5}~ \Omega\)
The average power dissipated by resistor (power is never dissipated by inductor or capacitor) of resistance π in any ac circuit with πππ current \(πΌ_{πππ }\) is given by,
\(πΌ_{πππ }=\dfrac{π_{πππ }}{π}=\dfrac{6}{\sqrt{5}} ~π΄\)
Power dissipated,
\(π=π_{πππ } πΌ_{ππ}~ πππ β‘\phi=π_{πππ } πΌ_{πππ }\dfrac{ π }{π}\)
\(π=6\times\dfrac{6}{\sqrt{5}}\times\dfrac{2}{\sqrt{5}}=\dfrac{72}{5}=14.4 ~π\)
Final Answer: Option (c)
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