Question

An electrical device draws 2kW power from ac mains voltage 223 V(rms). The current differs in phase by $\phi ={{\tan }^{-1}}\left( \dfrac{-3}{4} \right)$ as compared to the voltage. The resistance R in the circuit is
a) 15 $\Omega$
b) 20 $\Omega$
c) 25 $\Omega$
d) 30 $\Omega$

Answer 1

The power consumed by the device is the same as consumed by the device in the DC circuit. The voltage given is the rms value of the voltage and hence we can use the relation between the power and the voltage as that in the DC circuit. If the value of the rms voltage across the electrical device is (V) of impedance(Z), than the power dissipated across the electrical device is $P=\dfrac{{{V}^{2}}}{Z}Watts$. It is also given to us that the current lag in the circuit is $\phi ={{\tan }^{-1}}\left( \dfrac{-3}{4} \right)$. Therefore we will obtain the impedance in the device in terms of resistance of the device by using the are required relation from the above equation and substitute in the formula of power to obtain the resistance.

Complete step-by-step answer:
Let us say the above electrical device consists of capacitor, inductor and resistor. Let the resistance of the resistor be (R)capacitive reactance of the capacitor be ${{X}_{C}}$ and the inductive reactance of the inductor be ${{X}_{L}}$. Than current lags behind the voltage by angle $\phi$ which is given by,
$\begin{aligned} & \phi ={{\tan }^{-1}}\left( \dfrac{{{X}_{L}}-{{X}_{C}}}{R} \right) \\\ & \Rightarrow \tan \phi =\dfrac{{{X}_{L}}-{{X}_{C}}}{R}...(1) \\\ \end{aligned}$
In the question it is given to us that the current lags behind the voltage by,
$\begin{aligned} & \phi ={{\tan }^{-1}}\left( \dfrac{-3}{4} \right) \\\ & \Rightarrow \tan \phi =\dfrac{-3}{4}...(2) \\\ \end{aligned}$
Equating equation 1 and 2 we get,
$\begin{aligned} & \tan \phi =\dfrac{{{X}_{L}}-{{X}_{C}}}{R}=\dfrac{-3}{4}. \\\ & \Rightarrow {{X}_{L}}-{{X}_{C}}=\dfrac{-3R}{4} \\\ \end{aligned}$
The impedance(Z) in a circuit is equal to, $Z=\sqrt{{{R}^{2}}+{{({{X}_{L}}-{{X}_{C}})}^{2}}}$
Hence the Impedance in the circuit in terms of resistance is equal to,
$\begin{aligned} & Z=\sqrt{{{R}^{2}}+{{({{X}_{L}}-{{X}_{C}})}^{2}}} \\\ & \Rightarrow Z=\sqrt{{{R}^{2}}+{{\left( \dfrac{3}{4}R \right)}^{2}}} \\\ & \Rightarrow Z=\sqrt{\dfrac{25{{R}^{2}}}{16}}=\dfrac{5R}{4} \\\ \end{aligned}$
The relation between power dissipated across the impedance Z in a DC is given by,
$P={{I}^{2}}Z$ but from ohms law since$V=IZ$ power can also be written as
$P=\dfrac{{{V}^{2}}}{Z}$
Now since the value of voltage and power given to us are rms values of an ac circuit we can use the above equation to find the resistance in the circuit.
$\begin{aligned} & P=\dfrac{{{V}^{2}}}{Z} \\\ & Z=\dfrac{{{V}^{2}}}{P} \\\ \end{aligned}$
Power dissipated =2kW, V=223 V and Z=5R/4. Substituting this in the above equation,
$\begin{aligned} & Z=\dfrac{{{V}^{2}}}{P} \\\ & \Rightarrow \dfrac{5R}{4}=\dfrac{{{223}^{2}}}{2\times {{10}^{3}}} \\\ & \Rightarrow R=\dfrac{2{{(223)}^{2}}}{5\times {{10}^{3}}}\Omega =19.89\Omega \\\ \end{aligned}$
R= 20$\Omega$ approx.
Hence the answer of the above question is option b.

So, the correct answer is “Option b”.

Note: RMS (root mean square) value current or voltage in an ac circuit is that value of steady current or voltage in DC, which produces the same amount of heat on connecting to the source.
Let us say a resistor R is connected to ac source of angular frequency $\omega$ of emf $E={{E}_{\circ }}\operatorname{sin}\omega t$ and the instantaneous value of current in the circuit is given by $I={{I}_{\circ }}sin\omega t$ where ${{I}_{\circ }}=\dfrac{E}{R}A$ is the maximum value of current in a circuit in a complete cycle. The rms value of the current is given by ${{I}_{rms}}=\dfrac{{{I}_{\circ }}}{\sqrt{2}}A$.