Question

A circuit shown in the figure contains a box having either a capacitor or an inductor. The power factor of the circuit is 0.8, while current lags behind the voltage. Then find the inductance/capacitance of the box in S.I. unit. (take $\pi$ = 3.2) The circuit draws 1 ampere current.

Answer 1

2.5

Answer 2

The circuit consists of a resistor $R = 800 \, \Omega$, an external capacitor with reactance $X_C$, and a box containing either a capacitor or an inductor. These components are in series with an AC voltage source of frequency $f = 100 \, Hz$. The current in the circuit is $I = 1 \, A$. The voltage across the external capacitor is given as $V_C = 1000 \, V$.

The reactance of the external capacitor is $X_C = \frac{V_C}{I} = \frac{1000 \, V}{1 \, A} = 1000 \, \Omega$. The capacitive reactance is given by $X_C = \frac{1}{\omega C}$, where $\omega = 2\pi f$. Given $f = 100 \, Hz$ and $\pi = 3.2$, $\omega = 2 \times 3.2 \times 100 = 640 \, rad/s$. So, $1000 = \frac{1}{640 C}$, which gives $C = \frac{1}{640 \times 1000} = \frac{1}{640000} \, F$. The reactance of the external capacitor is $X_C = -1000 \, \Omega$.

The power factor of the circuit is given as $\cos \phi = 0.8$, and the current lags behind the voltage. In a series AC circuit, the power factor is given by $\cos \phi = \frac{R_{total}}{|Z|}$, where $R_{total}$ is the total resistance and $|Z|$ is the magnitude of the total impedance. Let the impedance of the box be $Z_{box} = R_{box} + jX_{box}$. The total impedance of the circuit is $Z = R + Z_{box} + Z_C = (R + R_{box}) + j(X_{box} + X_C)$. Since the box contains either a capacitor or an inductor, it is a purely reactive component, so $R_{box} = 0$. Thus, the total impedance is $Z = R + j(X_{box} + X_C)$. The total resistance is $R_{total} = R = 800 \, \Omega$. The magnitude of the total impedance is $|Z| = \sqrt{R^2 + (X_{box} + X_C)^2}$. The power factor is $\cos \phi = \frac{R}{|Z|} = 0.8$. So, $0.8 = \frac{800}{|Z|}$, which gives $|Z| = \frac{800}{0.8} = 1000 \, \Omega$.

Now, $|Z|^2 = R^2 + (X_{box} + X_C)^2$. $1000^2 = 800^2 + (X_{box} - 1000)^2$. $1000000 = 640000 + (X_{box} - 1000)^2$. $(X_{box} - 1000)^2 = 1000000 - 640000 = 360000$. $X_{box} - 1000 = \pm \sqrt{360000} = \pm 600$. So, $X_{box} = 1000 + 600 = 1600 \, \Omega$ or $X_{box} = 1000 - 600 = 400 \, \Omega$.

The phase angle $\phi$ is given by $\tan \phi = \frac{X_{total}}{R} = \frac{X_{box} + X_C}{R}$. Since the current lags behind the voltage, the phase angle $\phi$ is positive, which means the total reactance $X_{total} = X_{box} + X_C$ is positive.

Case 1: $X_{box} = 1600 \, \Omega$. Then $X_{total} = X_{box} + X_C = 1600 - 1000 = 600 \, \Omega$. Since $X_{total} = 600 > 0$, the current lags the voltage. This is consistent with the given information. If $X_{box} = 1600 \, \Omega$ is the reactance of the box, and the box contains either a capacitor or an inductor, then since the reactance is positive, the box must contain an inductor. For an inductor, $X_L = \omega L$. So, $1600 = 640 \times L$. $L = \frac{1600}{640} = \frac{160}{64} = \frac{10}{4} = 2.5 \, H$.

Case 2: $X_{box} = 400 \, \Omega$. Then $X_{total} = X_{box} + X_C = 400 - 1000 = -600 \, \Omega$. Since $X_{total} = -600 < 0$, the current leads the voltage. This contradicts the given information that the current lags behind the voltage. Therefore, this case is not possible.

Thus, the box contains an inductor with inductance $L = 2.5 \, H$.