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In series LCR circuit voltage drop across resistance is 8V, across inductor is 6V and across capacitor is 12V. then
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a
Voltage of the source will be leading current in the circuit
b
Voltage drop across each element will be less than the applied voltage
c
Power factor of circuit will be 4/3
d
None of these
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Correct option is D
Since cosθ=RZ=IRIZ=810=45 (also, cosθ can never be greater than 1)Hence (c) is wrong.Also IXC>IXL⇒XC>XL Current will be leading.In an LCR circuit,v=(vL−vC)2+vR2=(6−12)2+82 V=10; which is less than voltage drop across capacitor.Similar Questions
In the series LCR circuit as shown in figure, the heat developed in 80 seconds and amplitude of wattless current is:
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