Q.In Exercises 7.3 and 7.4, what is the net power absorbed by each circuit over a complete cycle. Explain your answer.

Concept and Intuition

In a pure AC circuit, instantaneous power oscillates, but the useful (average) power dissipated depends on how much the current is in phase with the voltage. If current lags/leads voltage by $\phi$, only the in-phase component $I_0\cos\phi$ contributes to the time-averaged power — this is captured by the power factor $\cos\phi$. The RMS values are used because they represent the equivalent DC values that would dissipate the same average power.

Step-by-Step Solution

1. Given $V=100\sin(100t)$ V $\Rightarrow V_0=100$ V, phase $=0$.
2. Given $I=100\sin\left(100t+\dfrac{\pi}{3}\right)$ mA $\Rightarrow I_0=100$ mA $=0.1$ A, and current leads voltage by $\phi=\pi/3=60^\circ$.
3. RMS values: $V_{rms}=\dfrac{V_0}{\sqrt2}=\dfrac{100}{\sqrt2}$ V, $I_{rms}=\dfrac{I_0}{\sqrt2}=\dfrac{0.1}{\sqrt2}$ A. …

Concept and Intuition

In an AC circuit with a phase difference $\phi$ between voltage and current (as happens with any reactive element), only the in-phase component of current does real work — hence the $\cos\phi$ (power factor) term. Using peak values, $P_{avg} = \dfrac{V_0 I_0}{2}\cos\phi$, equivalent to $V_{rms}I_{rms}\cos\phi$.

Step-by-Step Solution

1. From $V=50\sin(50t)$ V, $V_0 = 50$ V; from $I=50\sin(50t+\pi/4)$ mA, $I_0 = 50\ \mathrm{mA} = 0.05$ A, and phase difference $\phi=\pi/4$.
2. Average power: $P = \dfrac{V_0 I_0}{2}\cos\phi = \dfrac{50\times0.05}{2}\cos(45^\circ)$.
3. $\dfrac{50\times0.05}{2} = \dfrac{2.5}{2} = 1.25$. …

Concept and Intuition

In a series AC circuit with resistance $R$ and inductive reactance $X_L$, the voltage across $R$ and across $L$ are 90° out of phase, so they combine as vectors (phasors) at right angles. The impedance is the phasor sum:

The power factor is the cosine of the phase angle between current and applied voltage, given by $\cos\phi = R/Z$ (since the resistor is the only element that actually dissipates real power).

Step-by-Step Solution

1. $Z = \sqrt{R^2+X_L^2} = \sqrt{60^2+80^2} = \sqrt{3600+6400}=\sqrt{10000}=100\ \Omega$. …

Concept and Intuition

In a series LCR circuit, the power factor is $\cos\phi=\dfrac{R}{Z}$, where $Z=\sqrt{R^2+(X_L-X_C)^2}$ is the total impedance. The resistor is the only element that dissipates real power; the reactive elements only shift the phase, and the power factor tells us what fraction of the apparent power (VI) is actually real (dissipated) power.

Step-by-Step Solution

1. Net reactance: $X=X_L-X_C=2R-3R=-R$; its magnitude is $R$ (the circuit is net capacitive, but only the magnitude matters for the power factor).
2. Impedance: $Z=\sqrt{R^2+X^2}=\sqrt{R^2+R^2}=\sqrt{2R^2}=R\sqrt2$. …

Concept and Intuition

In a series R–L ac circuit, the resistance $R$ and inductive reactance $X_L$ combine as perpendicular sides of a right triangle whose hypotenuse is the impedance $Z = \sqrt{R^2+X_L^2}$. The phase angle $\phi$ by which current lags the applied voltage satisfies $\cos\phi = R/Z$ and $\tan\phi = X_L/R$. The power factor $\cos\phi$ tells us directly what fraction of $Z$ is resistive, which pins down the angle and hence the ratio of the two triangle legs.

Step-by-Step Solution

1. Given power factor $\cos\phi = R/Z = 0.5$, so $\phi = 60^\circ$.
2. In the impedance triangle, $\tan\phi = \dfrac{X_L}{R}$. …

Concept and Intuition

Power factor is $\cos\phi = \dfrac{R}{Z}$, where $Z=\sqrt{R^2+X^2}$ and $X$ is the net reactance (inductive minus capacitive). Adding a capacitor doesn't just add resistance to the circuit's opposition — it can partially cancel the inductive reactance, and here it happens to leave the net reactance magnitude unchanged.

Step-by-Step Solution

1. LR circuit: $X_L=R$ (given). Impedance $Z_1=\sqrt{R^2+X_L^2}=\sqrt{R^2+R^2}=R\sqrt2$. Power factor: $\cos\phi_1 = \dfrac{R}{Z_1}=\dfrac{R}{R\sqrt2}=\dfrac{1}{\sqrt2}$.
2. LCR circuit: capacitor with $X_C=2R$ added in series. Net reactance $=X_L-X_C = R-2R=-R$ (magnitude $R$). Impedance $Z_2=\sqrt{R^2+(X_L-X_C)^2}=\sqrt{R^2+R^2}=R\sqrt2$. …

Concept and Intuition

For an inductor on an AC source, $v=V_0\sin(\omega t)$ and $i=I_0\sin(\omega t - \pi/2)$ (current lags voltage by 90°). The instantaneous power is $p=vi=V_0 I_0 \sin(\omega t)\sin(\omega t-\pi/2)$. Using the product-to-sum trigonometric identity, this simplifies to a term containing $\cos(2\omega t)$ — i.e. the power oscillates at twice the angular frequency of the source. This is a general feature: the instantaneous power in any AC circuit (resistive, inductive, or capacitive) oscillates at twice the source frequency, since it's built from products of sinusoids of the same base frequency.

Step-by-Step Solution

1. Source frequency: $f = 50\ \text{Hz}$, so $\omega = 2\pi f$.
2. Voltage and current both vary as $\sin(\omega t + \text{phase})$.
3. Instantaneous power $p(t) = v(t)i(t)$ is a product of two sinusoids of the same frequency $\omega$. …