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Nax13
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A resistance 'R' ohm and inductance of 'L' henry are connected across 240V, 50Hz supply. Power distributed in the circuit is 300W, and the voltage across R is 100V. In order to improve the power factor to unity, the capacitor to be connected in series should have a value of __.
a) 43.7 uF b) 4.37 uF c) 437 uF d) 0.437 uF
Step 1) Using P=V/I, got I=1.25A
Step 2) Using R=VR/I, got R=80 ohm
Step 3) Now since total voltage is 240V, VL is 240-100=140V
Step 4) Using XL=VL/I, got XL=112 ohm.
Step 5) For power factor to be unity, XL=XC, which means that XC should also be equal to 112 ohm, which leads me to the answer that C is 28.4uF. (since XC=1/2πfC)
Where did I go wrong? Any help would be appreciated.
a) 43.7 uF b) 4.37 uF c) 437 uF d) 0.437 uF
Step 1) Using P=V/I, got I=1.25A
Step 2) Using R=VR/I, got R=80 ohm
Step 3) Now since total voltage is 240V, VL is 240-100=140V
Step 4) Using XL=VL/I, got XL=112 ohm.
Step 5) For power factor to be unity, XL=XC, which means that XC should also be equal to 112 ohm, which leads me to the answer that C is 28.4uF. (since XC=1/2πfC)
Where did I go wrong? Any help would be appreciated.
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