Question

In the circuit shown in the figure power factor of the box is $0.5$ and the power factor of the circuit is $\dfrac{{\sqrt 3 }}{2}$. The current is leading the voltage. Find the effective resistance (in ohms) of the box.

Answer 1

Hint: Capacitance opposes the change in voltage and serves to delay the increase or decrease of voltage across the capacitor. This causes the voltage to lag behind the current in a capacitive circuit. When L or C is present in an ac circuit, energy is required to build up a magnetic field around L or an electric field in C. This energy comes from the source.

Complete step by step answer:
Given, power factor of the box is given by, $\cos {\phi _{box}} = 0.5$
Then, ${\phi _{box}} = {\cos ^{ - 1}}(0.5)$
$ \Rightarrow {\phi _{box}} = {60^0}$
Similarly, power factor of the circuit is given by, $\cos {\phi _{circuit}} = \dfrac{{\sqrt 3 }}{2}$
Then, ${\phi _{circuit}} = {\cos ^{ - 1}}(\dfrac{{\sqrt 3 }}{2})$
$ \Rightarrow {\phi _{circuit}} = {30^0}$
Given, resistance in the circuit $R = 10\Omega $
Here we need to find the resistance to the box.
Let ${R_{box}}$ be the effective resistance of the box.
It is given in the question that the current is leading the voltage. From this, we can say that the given circuit is a purely capacitive circuit.
Then$\tan {\phi _{box}} = \dfrac{{{X_c}}}{{{R_{box}}}}$
Where ${X_C}$ is the reactive capacitance of the circuit.
We have, ${\phi _{box}} = {60^0}$and $\tan {60^0} = \sqrt 3 $ then
Above equation becomes, $\sqrt 3 = \dfrac{{{X_c}}}{{{R_{box}}}}$
$ \Rightarrow {X_C} = \sqrt 3 {R_{box}}$ ………………..(1)
Similarly,
$ \Rightarrow \tan {\phi _{circuit}} = \dfrac{{{X_c}}}{{{R_{circuit}} + {R_{box}}}}$
We have, ${\phi _{box}} = {30^0}$and $\tan {30^0} = \dfrac{1}{{\sqrt 3 }}$
Then the above equation becomes,
$ \Rightarrow \dfrac{1}{{\sqrt 3 }} = \dfrac{{{X_c}}}{{{R_{circuit}} + {R_{box}}}}$
$ \Rightarrow {X_c} = \dfrac{{{R_{circuit}} + {R_{box}}}}{{\sqrt 3 }}$
We can substitute the values in the equation we get,
$ \Rightarrow {X_c} = \dfrac{{10 + {R_{box}}}}{{\sqrt 3 }}$ ……………………(2)
Compare equation (1) and (2) we get,
$ \Rightarrow \sqrt 3 {R_{box}} = \dfrac{{10 + {R_{box}}}}{{\sqrt 3 }}$
Simplifying the above equation we get,
$ \Rightarrow \sqrt 3 \times \sqrt 3 {R_{box}} = 10 + {R_{box}}$
$ \Rightarrow 3{R_{box}} = 10 + {R_{box}}$
$\therefore {R_{box}} = 5\Omega $

Therefore, the effective resistance (in ohms) of the box is $5\Omega $.

Note:
Consider a circuit containing an inductor, capacitor and resistor connected in series across an alternating source of voltage $V$ or emf $\varepsilon $ .

Let the source supplies a sinusoidal voltage which is given by,
$V = {V_0}\sin \omega t$
Where, ${V_0}$ is the peak value of voltage $\omega $ is the angular frequency and $t$ is the time period.
Let $q$ be the charge on the capacitor and $I$ be the current in the circuit at any instant of time $t$.
Let${V_R},{V_L},{V_C}$ represent the voltage across the resistor, inductor, and capacitor respectively.
Then, voltage across resistor, ${V_R} = {i_0}R$
Voltage across inductor, ${V_L} = {i_0}{X_L}$
Voltage across capacitor, ${V_C} = {i_0}{X_C}$
Where, ${i_0}$ is the peak value of current, ${X_C}$ is capacitive reactance, ${X_L}$ is the inductive reactance and $R$ is the resistance of the resistor.
Then net voltage or emf is given by, $V$ or $\varepsilon = \sqrt {\left( {{V_R}^2 + {{\left( {{V_L} - {V_C}} \right)}^2}} \right)} $