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  • In series LCR circuit voltage drop across resistance is 8 volt and across capacitor is 12 volt. Then :Option: 1 Voltage of the source will be leading

In series LCR circuit voltage drop across resistance is 8 volt and across capacitor is 12 volt. Then :

Option: 1

Voltage of the source will be leading current in the circuit


Option: 2

Voltage drop across each element will be less than the applied voltage


Option: 3

power factor of circuit will be 4/3


Option: 4

None of these


Answers (1)

best_answer

\mathrm{\text { Since, } \cos \theta=\frac{\mathrm{R}}{\mathrm{Z}}=\frac{\mathrm{IR}}{\mathrm{IZ}}=\frac{8}{10}=\frac{4}{5}}

(\mathrm{\cos\theta} can never be greater than 1)

Also, \mathrm{\mid x_C>1 x_L \quad \Rightarrow x_C>x_L}

Current will be leading In a LCR circuit

\mathrm{\mathrm{V}=\sqrt{\left(\mathrm{V}_{\mathrm{L}}-\mathrm{V}_{\mathrm{C}}\right)^2} \quad=\sqrt{(6-12)^2+8^2}}

V = 10 ; which is less than voltage drop across capacitor

Posted by

avinash.dongre

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