Question

In series L-C-R circuit voltage drop across the resistance is 8V, across inductor is 6V and across capacitor is 12V. Then:
a. Voltage of the source will be leading.
b. Voltage drop across each element will be less than the applied voltage.
c. Power factor of the circuit is$\dfrac{3}{4}$.
d. None of the above.

Answer 1

Hint: By using the given values first we have to calculate all the parameters that are given in the options. This will give us a fair idea of which one will be the correct option. The formula of the resultant voltage of the LCR circuit can be used to calculate the required answer.

Formula used:
The formula of the resultant voltage is given by $ \Rightarrow V = \sqrt {{V_R}^2 + {{\left( {{V_L} - {V_C}} \right)}^2}} $ where ${V_R}$is the voltage across the resistor, ${V_C}$ is the voltage across the capacitor, ${V_L}$is the voltage across the inductor and V is the voltage for the LCR circuit.

Complete step by step answer:
It is given in the problem that${V_R} = 8V$, ${V_C} = 12V$and${V_L} = 6V$. In a series LCR circuit, the applied voltage can be given as$ \Rightarrow V = \sqrt {{V_R}^2 + {{\left( {{V_C} - {V_L}} \right)}^2}} $.
Please note that when we use the above equation, the output of the reactive voltage must always be a positive value, that is we have to subtract the smaller value from the larger one which will give us a positive number which will, in turn, be added to ${V_R}$ thus we can use both${V_L} - {V_C}$or${V_C} - {V_L}$.
Now replace the values of${V_R}$ ,${V_C}$, and ${V_L}$ in the above equation we get,
$ \Rightarrow V = \sqrt {{8^2} + {{\left( {12 - 6} \right)}^2}} $
$ \Rightarrow V = \sqrt {64 + 36} $
$ \Rightarrow V = \sqrt {100} $
$ \Rightarrow V = 10$
Thus, $V = 10V$.
Now let us calculate the power factor,
\[P \cdot F = \cos \phi = \dfrac{R}{Z}\],
Where P.F is the power factor, R is the reactance of the circuit, and Z is the impedance of the circuit.
Now multiplying current I in the above equation we get,
$ \Rightarrow \cos \phi = \dfrac{{IR}}{{IZ}}$
$ \Rightarrow \cos \phi = \dfrac{{{V_R}}}{V}$
Now putting the values of ${V_R}$ and V in the equation,
$ \Rightarrow \cos \phi = \dfrac{8}{{10}}$
$ \Rightarrow \cos \phi = \dfrac{4}{5}$
Also, it is given in the problem that.
$ \Rightarrow {V_L} < {V_C}$
So, we get.
$ \Rightarrow I({X_L}) < I\left( {{X_C}} \right)$.
$ \Rightarrow {X_L} < {X_C}$

Hence, the correct answer is option (D).

Additional Information:
There is a difference between the resistance and impedance, although both have the same units. Resistance is simply defined as the opposition to the flow of electric current in the circuit whereas Impedance is opposition to the flow of AC current because of any three components that are resistive, inductive or capacitive. It is a combination of both resistance and reactance in a circuit. Resistance occurs in both AC and DC circuits, whereas Impedance takes place only in an AC circuit. Resistance is the contribution of the resistive element in the circuit, whereas the contribution of both resistance and reactance forms Impedance.

Note: Remember that if ${V_c} > {V_L}$ then current will always lead the voltage. Always remember that the power factor is the ratio of the real power that is used to do work and the apparent power that is supplied to the circuit. The power factor can get values in the range from 0 to 1. When all the power is reactive power then the power with no real power (usually inductive load) the power factor is 0.