Unit 04: Alternating Current • Chapter 07
Power Factor, Wattless Current & The Transformer
Real vs apparent power, power factor correction, quadrature wattless current, mutual flux induction, and laminated core losses.
Socho ek fizzy cold drink ke glass ko: glass mein jo liquid hai woh asal drink hai jise aap peete hain (Active / Real Power: $P = V_{\text{rms}}I_{\text{rms}}\cos\phi$), lekin upar jo jhaag (foam) banta hai woh glass ki jagah toh gher leta hai par pyaas nahi bujhata (Wattless / Reactive Current: $I_{\text{rms}}\sin\phi$). Poora glass foam samet Apparent Power ($V_{\text{rms}}I_{\text{rms}}$) hai. Agar jhaag zyada ho, toh electricity company ke motey taar bekar mein current carry karte hain bina koi useful work kiye! Aur jab is power ko hazaron kilometer door bhejna hota hai, toh raste mein taaron ki garmi ($I^2R$) se bachne ke liye hum Transformer use karte hain: voltage ko hazaron guna badha kar current ko lagbhag zero kar diya jaata hai, taaki energy bina loss ke safely transport ho sake!
1. AC Power & Wattless Current
Average power dissipated over a complete cycle:
Pure $L$ or $C$: $\phi = 90^\circ \implies \cos 90^\circ = 0 \implies P_{\text{avg}} = 0$. Resonance / Pure $R$: $\phi = 0^\circ \implies \cos 0^\circ = 1$ (Maximum power).
2. Ideal Transformer Equations
Mutual flux linkage ratio with 100% efficiency:
Step-up Transformer: $k > 1$ ($N_s > N_p, V_s > V_p, I_s < I_p$). Step-down Transformer: $k < 1$ ($N_s < N_p, V_s < V_p, I_s > I_p$).
Average Power in AC Circuit & The Power Factor
1. Let applied instantaneous alternating voltage and resulting current with phase difference $\phi$ be:
2. Instantaneous power delivered to the circuit: $p = v \times i$:
3. Using trigonometric identities $\sin^2\omega t = \frac{1 - \cos 2\omega t}{2}$ and $\sin\omega t\cos\omega t = \frac{\sin 2\omega t}{2}$:
4. Average power over a complete cycle of time period $T$ ($P_{\text{avg}} = \frac{1}{T}\int_0^T p \, dt$):
• The time averages over a full cycle are: $\langle 1 \rangle = 1$, $\langle \cos 2\omega t \rangle = 0$, and $\langle \sin 2\omega t \rangle = 0$.
Resolution of Current: Active Component vs Wattless Component
Fig 7.3A: Resolution of current phasor $\vec{I}_{\text{rms}}$ into active power component ($I_{\text{rms}}\cos\phi$) and wattless component ($I_{\text{rms}}\sin\phi$).
1. In an AC circuit where current lags voltage by phase angle $\phi$, resolve $\vec{I}_{\text{rms}}$ into two orthogonal components:
• Active Component: $I_{\text{active}} = I_{\text{rms}}\cos\phi$ along $\vec{V}_{\text{rms}}$ (in-phase, phase difference $= 0$).
• Wattless Component: $I_{\text{wattless}} = I_{\text{rms}}\sin\phi$ perpendicular to $\vec{V}_{\text{rms}}$ (in quadrature, phase difference $= 90^\circ$).
2. Definition: The component of AC current which consumes zero average power over a complete cycle ($P_2 = 0$) is known as the Wattless Current ($I_{\text{rms}}\sin\phi$).
Transformer Principle, Formulation of Turns Ratio & Energy Losses
1. Principle: Works on the principle of Mutual Induction—whenever current in the primary coil changes, magnetic flux linked with the secondary coil changes, inducing an alternating EMF across it.
2. Let an alternating voltage $V_p$ be applied across the primary coil of $N_p$ turns. Let the common magnetic flux linked per turn of the soft iron core be $\Phi$.
3. Induced EMF in the primary coil (by Faraday-Lenz law):
4. The same mutual flux $\Phi$ threads each of the $N_s$ turns of the secondary coil. Induced EMF in the secondary coil:
5. Assuming ideal conditions (negligible winding resistance and zero flux leakage, so $\mathcal{E}_p = V_p$ and $\mathcal{E}_s = V_s$):
6. For an ideal transformer with zero power loss ($P_{\text{input}} = P_{\text{output}} \implies V_p I_p = V_s I_s$):
Four Major Energy Losses in Practical Transformers & Their Minimization:
• 1. Copper Loss ($I^2R$): Joule heat in copper windings. Remedy: Use thick copper wire with low resistance for high-current coils.
• 2. Eddy Current Loss: Circulating induced currents in the solid iron core. Remedy: Use a laminated core composed of thin sheets insulated by varnish.
• 3. Hysteresis Loss: Energy wasted in repeated cyclic magnetization and demagnetization of the core. Remedy: Use soft iron which has a very narrow hysteresis loop.
• 4. Flux Leakage: All flux produced by primary does not thread secondary. Remedy: Wind secondary directly over primary in a coaxial arrangement.
Trap 1: Connecting a Transformer to a DC Battery Source
*Board Question:* What happens if a $12\text{ V}$ DC battery is connected across the primary of a step-up transformer?
• DC produces a constant, steady magnetic flux ($d\Phi/dt = 0$).
• Secondary voltage is strictly ZERO ($V_s = 0$) because induction requires changing flux!
• In primary, back-EMF is zero ($X_L = 0$); only tiny ohmic resistance $R_p$ opposes current. A massive direct current flows, and the primary coil burns out immediately!
Trap 2: High Voltage Long-Distance Transmission Logic
*Why transmit power at $400{,}000\text{ V}$ instead of $220\text{ V}$?*
Power to be transmitted is constant: $P = V I \implies I = P/V$.
Heat loss in transmission cables of resistance $R_c$:
$$P_{\text{loss}} = I^2 R_c = \left(\frac{P}{V}\right)^2 R_c = \frac{P^2 R_c}{\mathbf{V^2}} \propto \frac{1}{V^2}$$
Stepping up voltage by a factor of $100$ reduces line power losses by a factor of $10{,}000$ ($100^2$)!
Trap 3: Conservation of Energy in Step-Up Transformers
Does a step-up transformer violate Conservation of Energy because $V_s > V_p$?
NO! When voltage increases ($V_s > V_p$), the current decreases in the exact same proportion ($I_s < I_p$).
Power output can never exceed power input: $P_s \le P_p$. Voltage is stepped up at the expense of current!
Trap 4: Frequency Invariance through a Transformer
A transformer changes voltage and current, but the frequency of the AC supply remains strictly UNCHANGED ($f_s = f_p$). If input is $50\text{ Hz}$, output is always $50\text{ Hz}$.
Problem: A transformer has 200 turns in the primary coil and 800 turns in the secondary coil. The primary is connected to an AC mains supply of $220\text{ V}, 50\text{ Hz}$. If a load of resistance $440\text{ }\Omega$ is connected across the secondary, find (i) secondary voltage, (ii) secondary current, (iii) primary current, and (iv) input power, assuming an ideal transformer.
Solution:
1. Given: $N_p = 200$, $N_s = 800$, $V_p = 220\text{ V}$, $R_L = 440\text{ }\Omega$.
Transformation ratio $k = \frac{N_s}{N_p} = \frac{800}{200} = 4$ (Step-Up).
2. Secondary current ($I_s$):
3. Primary current ($I_p$):
4. Input power:
Problem: In a series circuit, an alternating voltage $v = 200\sqrt{2}\sin(100t)\text{ V}$ produces a current $i = 5\sqrt{2}\sin(100t - \frac{\pi}{3})\text{ A}$. Calculate (i) peak and RMS voltage and current, (ii) power factor of the circuit, (iii) average active power consumed, and (iv) magnitude of the wattless current.
Solution:
1. From given equations: $V_0 = 200\sqrt{2}\text{ V} \implies V_{\text{rms}} = 200\text{ V}$.
$I_0 = 5\sqrt{2}\text{ A} \implies I_{\text{rms}} = 5\text{ A}$. Phase angle $\phi = \frac{\pi}{3} = 60^\circ$.
2. Power factor:
3. Average real power consumed:
4. Wattless current component:
Power Factor
$\cos\phi = \frac{R}{Z}$
Wattless Current
$I_w = I_{\text{rms}}\sin\phi$ ($P=0$)
Turns Ratio
$k = \frac{V_s}{V_p} = \frac{N_s}{N_p} = \frac{I_p}{I_s}$
Laminated Core
Cuts Eddy Current Loss