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CBSE Class 12 Physics 2023 Delhi Set 1 Paper

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Question : 27 of 35
Marks: +1, -0
An alternating voltage of 220 V220\ \mathrm{V} is applied across a device XX. A current of 0.22 A flows in the circuit and it lags behind the applied voltage in phase by π2\frac{\pi}{2} radian. When the same voltage is applied across another device YY, the current in the circuit remains the same and it is in phase with the applied voltage.
(i) Name the devices XX and YY and,
(ii) Calculate the current flowing in the circuit when the same voltage is applied across the series combination of XX and YY.
(i) Since, current lags behind the voltage, XX is an inductor.
Since, current and voltage remain in phase, YY is a resistor.
(ii) Since, XL  =  VI  =  2200.22X_L\;=\;\frac{V}{I}\;=\;\frac{220}{0.22}
  =1000 Ω\;=1000\ \Omega
And, R  =  2200.22R\;=\;\frac{220}{0.22}
  =1000 Ω\;=1000\ \Omega
Z  =R2+XL2Z\;=\sqrt{R^2+X_L^2}
  =(1000)2+(1000)2\;=\sqrt{(1000)^2+(1000)^2}
  =10002 Ω\;=1000\sqrt{2}\ \Omega
So, the required current =  220Z  =  22010002=0.155 A=\;\frac{220}{Z}\;=\;\frac{220}{1000\sqrt{2}}=0.155\ \mathrm{A}
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