Question

An electric device draws 2 kW power from ac mains voltage 223 V (rms). The current lags in phase by \[\phi = {\tan ^{ - 1}}\left( { - \dfrac{3}{4}} \right)\] as compared to voltage. The resistance R in the circuit is:
A. \[15\,\Omega \]
B. \[20\,\Omega \]
C. \[25\,\Omega \]
D. \[30\,\Omega \]

Answer 1

Hint: Compare the given phase angle with the expression for the phase angle in LCR circuit and determine the term \[{X_L} - {X_C}\] in terms of resistance R. Then determine the impedance in terms of R. Then use the relation between rms voltage and rms current and express the current. Use the relation between rms current, impedance and power to calculate the resistance of the circuit.

Formula used:
Phase angle, \[\tan \phi = \dfrac{{{X_L} - {X_C}}}{R}\]
Here, \[{X_L}\] is the inductive reactance, \[{X_C}\] is the capacitive reactance and R is the resistance.
Impedance, \[Z = \sqrt {{R^2} + {{\left( {{X_L} - {X_C}} \right)}^2}} \]
The rms current in the circuit is given as,
\[{I_{rms}} = \dfrac{{{V_{rms}}}}{Z}\]
Here, \[{V_{rms}}\] is the rms voltage.

Complete step by step answer:
The circuit has three components that are resistor, inductor and capacitor. We know that in series LCR circuit, the phase angle is given as,
\[\tan \phi = \dfrac{{{X_L} - {X_C}}}{R}\]
\[ \Rightarrow \phi = {\tan ^{ - 1}}\left( {\dfrac{{{X_L} - {X_C}}}{R}} \right)\] …... (1)
Here, \[{X_L}\] is the inductive reactance, \[{X_C}\] is the capacitive reactance and R is the resistance.

We have given the phase angle,
\[\phi = {\tan ^{ - 1}}\left( { - \dfrac{3}{4}} \right)\] …… (2)

Comparing equation (1) and (2), we get,
\[\dfrac{{{X_L} - {X_C}}}{R} = - \dfrac{3}{4}\]
\[ \Rightarrow {X_L} - {X_C} = - \dfrac{3}{4}R\] …… (3)

We know that the impedance of the AC circuit is expressed as,
\[Z = \sqrt {{R^2} + {{\left( {{X_L} - {X_C}} \right)}^2}} \]
Substituting \[{X_L} - {X_C} = - \dfrac{3}{4}R\] in the above equation, we get,
\[Z = \sqrt {{R^2} + {{\left( { - \dfrac{3}{4}R} \right)}^2}} \]
\[ \Rightarrow Z = \sqrt {{R^2} + \dfrac{9}{{16}}{R^2}} \]
\[ \Rightarrow Z = R\sqrt {1 + \dfrac{9}{{16}}} \]
\[ \Rightarrow Z = R\sqrt {\dfrac{{25}}{{16}}} \]
\[ \Rightarrow Z = \dfrac{5}{4}R\]

We know that the rms current in the circuit is given as,
\[{I_{rms}} = \dfrac{{{V_{rms}}}}{Z}\]
Substituting \[\dfrac{5}{4}R\] for Z and 223 V for \[{V_{rms}}\] in the above equation, we get,
\[{I_{rms}} = \dfrac{{223}}{{\dfrac{5}{4}R}}\]
\[ \Rightarrow {I_{rms}} = \dfrac{{178.4}}{R}\] …… (4)

We have the expression for the power in AC circuit is,
\[P = I_{rms}^2Z\]
Substituting \[P = 2 \times {10^3}\,{\text{W}}\] and \[{I_{rms}} = \dfrac{{178.4}}{R}\] in the above equation, we get,
\[2 \times {10^3} = {\left( {\dfrac{{178.4}}{R}} \right)^2} \times \dfrac{5}{4}R\]
\[ \Rightarrow 2 \times {10^3} = \dfrac{{4 \times {{10}^4}}}{R}\]
\[ \Rightarrow R = \dfrac{{4 \times {{10}^4}}}{{2 \times {{10}^3}}}\]
\[ \therefore R = 20\,\Omega \]

So, the correct answer is option B.

Note: If the reactance in the circuit is only due to the resistance, then the power is expressed as, \[P = {I^2}R\]. But if the circuit has an inductor and capacitor, then the power is expressed as, \[P = I_{rms}^2Z\], where, Z is the impedance of the circuit. The unit of resistance, inductive reactance and capacitive inductance is the same and it is ohm.