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
Correct Answer: D
Solution :
[d] Since, \[\cos \theta =\frac{R}{Z}=\frac{IR}{IZ}=\frac{8}{10}=\frac{4}{5}\] (\[cos\theta \] can never be greater than 1) Also, \[I{{x}_{C}}>I{{x}_{L}}\Rightarrow {{x}_{C}}>{{x}_{L}}\] Current will be leading In a LCR circuit \[V=\sqrt{{{({{V}_{L}}-{{V}_{C}})}^{2}}+V_{R}^{2}}\] \[=\sqrt{{{(6-12)}^{2}}+{{8}^{2}}}\] V = 10; which is less than voltage drop across capacitor.
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