Question

A circuit $\text{A}$ takes a current of $3\text{A}$ at a power factor of $0.6$ lagging when connected to a $115\text{V},50\text{Hz}$ supply. Another circuit $\text{B}$ takes a current of $5\text{A}$ at a power factor of $0.707$ leading when connected to the same supply. If the two circuits are connected in series across a $230\text{V},50\text{Hz}$ supply. Calculate the power consumed in the circuit.

• A. $125\text{W}$
• B. $555.5\text{W}$
• C. $1187.5\text{W}$
• D. $1525.2\text{W}$

Answer 1

The correct option is C $1187.5\text{W}$

The impedance of circuit $\text{A}$ is given by,

$Z_A=\frac{E_0}{I_A}=\frac{115}{3}=38.33\Omega$

Current is lagging in the circuit $\text{A}$. Therefore, the circuit $\text{A}$ is a combination of a resistor and an inductor.

As,
$\cos \varphi =\frac{R}{Z}$ $\Rightarrow 0.6=\frac{R_A}{38.33}$

$\therefore R_A=23\Omega$

For an inductive circuit:

$Z_A=\sqrt{R_A^2+(X_L{)}_A^2}$

$(X_L{)}_A=\sqrt{Z_A^2-R_A^2}=\sqrt{{38.33}^2-{23}^2}=30.67\Omega$

Similarly, the impedance of circuit $\text{B}$ is,

$Z_B=\frac{115}{5}=23\Omega$

Current is leading in circuit $\text{B}$, therefore, the circuit $\text{B}$ is a combination of a resistor and a capacitor.

Now, $\cos \varphi =\frac{R_B}{Z_B}$

$R_B=0.707\times 23=16.26\Omega$

For a capacitive circuit:

$Z_B=\sqrt{(X_C{)}_B^2+R_B^2}$

$(X_C{)}_B=\sqrt{{23}^2-(16.26{)}^2}=16.26\Omega$

When circuits $\text{A}$ and $\text{B}$ are connected in the series:

$R=R_A+R_B=23+16.26=39.26\Omega$

$X_L-X_C=30.67-16.26=14.41\Omega$

$\therefore$ The impedance of the circuit is,

$Z=\sqrt{R^2+(X_L-X_C{)}^2}$

$=\sqrt{(39.26{)}^2+(14.41{)}^2}$

$\therefore Z=41.82\Omega$

The average power consumed by the circuit is,

$P_{avg}=V_{rms}{i}_{rms}\cos \varphi$

$=V_{rms}\left(\frac{V_{rms}}{Z}\right)\left(\frac{R}{Z}\right)$

$=230\times \frac{230}{41.82}\times \frac{39.26}{41.82}$

$\therefore P_{avg}=1187.5\text{W}$

Hence, $\left(\text{C}\right)$ is the correct answer.