Question

An inductor $20\,mH$ , a capacitor $100\mu F$ and a resistor of $50\Omega$ are connected in series across a source of emf, $V = 10\sin 314t$ . The power loss in the circuit is?
A. $2.74W$
B. $0.79W$
C. $1.13W$
D. $0.43W$

Answer 1

Here, in the example, we have to calculate the power loss in the circuit. Therefore, the solution will contain four steps. We will calculate inductive reactance and then capacitive reactance and using these we will calculate the impedance in the circuit. Using these, we calculate the power loss in the circuit.

Formula used:
The formula of the inductive reactance is
${X_L} = \omega L$
Here, ${X_L}$ is the inductive reactance, $\omega$ is the angular frequency and $L$ is the inductance.
The formula of the capacitive reactance is
${X_C} = \dfrac{1}{{\omega C}}$
${X_C}$ is the capacitive reactance and $C$ is the capacitance.
The formula of the impedance is
$Z = \sqrt {{R^2} + {{\left( {{X_C} - {X_L}} \right)}^2}}$
Now, the formula of the power loss is given by
$P = \dfrac{{V_0^2R}}{{2{Z^2}}}$
Here, $P$ is the power loss and ${V_0}$ is the voltage.

Complete step by step answer:
It is given in the question that the inductance of the inductor is,
$L = 20mH = 20 \times {10^{ - 3}}H$
Also, the resistance of the resistor is,
$R = 50\Omega$
The capacitance of the capacitor is,
$C = 100\mu F = 100 \times {10^{ - 6}}F$
If we consider a plane in which $R$ is connected in horizontal axis and ${X_L} - {X_C}$ is in the vertical direction and $Z$ makes an angle $\theta$ with the horizontal axis.
Now, the emf in the combination is $V = 10\sin 314t$
Comparing the above equation with $V = {V_0}\sin \omega t$ , we get
${V_0} = 10$ and $\omega = 314$
Now, the inductive reactance is defined as the resistance offered by the inductor to the flow of current. The formula used for calculating inductive reactance is given by
${X_L} = \omega L$
$\Rightarrow \,{X_L} = 314 \times 20 \times {10^{ - 3}}$
$\Rightarrow \,{X_L} = 6280 \times {10^{ - 3}}$
$\Rightarrow \,{X_L} = 6.280\Omega$
Also, the capacitive reactance is the resistance offered by the capacitor to the flow of current. The formula used for calculating capacitive reactance is given by
${X_C} = \dfrac{1}{{\omega C}}$
$\Rightarrow \,{X_C} = \dfrac{1}{{314 \times 100 \times {{10}^{ - 6}}}}$
$\Rightarrow \,{X_C} = . = 0.00003185 \times {10^6}$
$\Rightarrow \,{X_C} = 31.85\Omega$
Now, the impedance in the circuit is given by
$Z = \sqrt {{R^2} + {{\left( {{X_C} - {X_L}} \right)}^2}}$
$\Rightarrow \,Z = \sqrt {{{\left( {50} \right)}^2} + {{\left( {31.8 - 6.28} \right)}^2}}$
$\Rightarrow \,Z = \sqrt {2500 + {{\left( {25.52} \right)}^2}}$
$\Rightarrow \,Z = \sqrt {2500 + 651.27}$
$\Rightarrow \,Z = \sqrt {3151.27}$
$\Rightarrow \,Z = 56.13$
Now, the power loss in the circuit can be calculated by the given formula
$P = \dfrac{{V_0^2R}}{{2{Z^2}}}$
$\Rightarrow \,P = \dfrac{{{{\left( {10} \right)}^2} \times 50}}{{2 \times {{\left( {56.13} \right)}^2}}}$
$\Rightarrow \,P = \dfrac{{100 \times 50}}{{2 \times 3150.6}}$
$\Rightarrow \,P = \dfrac{{5000}}{{6301.2}}$
$\therefore \,P = 0.79W$
Therefore, the power loss in the series circuit of capacitor, inductor and resistor is $0.79W$ .

Hence, option B is the correct option.

Note: In the above example, $Z$ is the impedance that is the measure of total opposition to the current. It consists of both the resistance and the reactance. Now, the power lost in the circuit is the cause of the resistance or friction.Alternating current can be defined as a current that changes its magnitude and polarity at a regular interval of time. It can also be defined as an electrical current which repeatedly changes or reverses its direction opposite to that of Direct Current or DC which always flows in a single direction