Q.(a) For circuits used for transporting electric power, a low power factor implies large power loss in transmission. Explain.

(a) Low power factor → large transmission loss

The power factor $\cos \phi$ is the ratio of true power to apparent power:

$P = V_{\text{rms}} I_{\text{rms}} \cos \phi$.

For a given transmitted power $P$ and fixed voltage $V$, a low $\cos \phi$ forces a larger current $I$ to flow:

$I = \dfrac{P}{V \cos \phi}$.

Since transmission lines have resistance $R$, the power lost as heat is $I^2 R$. A larger $I$ means much larger ($\propto 1/\cos^2 \phi$) ohmic loss. Hence low power factor is economically wasteful.

(b) Capacitor improves power factor

Most industrial loads are inductive (motors, transformers), drawing a lagging current. The current phasor $I$ lags voltage $V$ by angle $\phi$, as shown in Figure 7.15 below. Its wattless component $I_q$ (horizontal, to the right) contributes no real power but increases the total current.

Adding a capacitor of appropriate capacitance draws a leading wattless current $I'_q$ (horizontal, to the left in the figure). This cancels part of the lagging $I_q$, reducing the net quadrature component. The resultant current $I$ becomes more in phase with $V$ — $\phi$ decreases, $\cos \phi$ increases — without changing the real power delivered.

(a) Why low power factor causes large transmission losses

1. What power factor actually means. In an AC circuit, the power factor $\cos\phi$ is the cosine of the phase angle between voltage and current. Only the component of current that is in phase with the voltage — called the active or power component $I_p = I\cos\phi$ — delivers real power $P = V_{\text{rms}} I_{\text{rms}} \cos\phi$. The perpendicular component $I_q = I\sin\phi$ is wattless; it sloshes energy back and forth between source and load but does zero net work.

2. The transmission line doesn't care about phase. The wires connecting the power station to the consumer have resistance $R$. The power lost as heat in these lines is $P_{\text{loss}} = I_{\text{rms}}^2 R$, where $I_{\text{rms}}$ is the total rms current — both the useful part and the useless part. The utility company must deliver a fixed real power $P$ to the customer. From $P = V_{\text{rms}} I_{\text{rms}} \cos\phi$, if $\cos\phi$ is small, the required $I_{\text{rms}}$ must be large to maintain the same $P$.

3. The consequence. A larger total current means dramatically larger line losses because loss scales as $I^2$. For example, if $\cos\phi$ drops from 1.0 to 0.5, the current must double to deliver the same real power, and the $I^2R$ loss quadruples. This is why power companies penalise industrial consumers with low power factors — the wasted heat in transmission lines is real money.

(b) How a capacitor improves power factor

4. The geometry of the phasor diagram. Look at Figure 7.15 below. The applied voltage $\vec{V}$ points vertically upward. The load current $\vec{I}$ lags behind $\vec{V}$ by angle $\phi$ (typical for inductive loads like motors and transformers). This current is resolved into:
   • $I_p = I\cos\phi$ — vertical, in phase with $V$ (the power component)
   • $I_q = I\sin\phi$ — horizontal, to the right (the lagging wattless component)

5. What a capacitor does. A pure capacitor draws a current that leads the voltage by $90^\circ$. In the phasor diagram, this leading current $\vec{I}'_q$ points horizontally to the left — exactly opposite to the lagging $I_q$. When we connect a capacitor in parallel with the load, the total line current becomes the vector sum of the load current and the capacitor current.

6. Cancelling the reactive component. If we choose the capacitor such that $I'_q = I_q$, the horizontal components cancel completely. The net line current becomes purely vertical — in phase with the voltage. The phase angle becomes zero, and $\cos\phi = 1$. Even partial cancellation (a smaller capacitor) reduces $\phi$ and improves the power factor.

7. Why this reduces transmission losses. After adding the capacitor, the total line current $I_{\text{new}}$ is smaller than the original $I$ for the same delivered real power $P$. Since $P_{\text{loss}} = I_{\text{new}}^2 R$, the reduction in current directly reduces the heat wasted in the transmission lines. The capacitor itself is nearly lossless (ideal capacitors dissipate negligible power), so the improvement comes essentially for free.

Concept: Capacitive Reactance

Method: Power Factor Analysis Using Phasor Diagrams

Why this method? Power factor problems are best understood by visualizing the phase relationship between voltage and current. A phasor diagram makes the role of capacitance immediately clear.

(a) Low Power Factor → Large Transmission Loss

Step 1: Recall the power factor definition

Power factor ($\cos \phi$) is the cosine of the phase angle between voltage and current in an AC circuit.

Step 2: Understand the current implication

For a given real power $P$ transmitted:

If $\cos \phi$ is low (say 0.5), the current $I_{\text{rms}}$ must be larger to deliver the same $P$.

Step 3: Connect to transmission loss

Power loss in transmission lines (resistive heating) is:

Since $I_{\text{rms}}$ is inversely proportional to $\cos \phi$, a low power factor means much larger $I^2R$ losses.

Key result: Low $\cos \phi$ → high current → large $I^2R$ loss in transmission lines.

(b) Capacitor Improves Power Factor

Step 1: Identify the cause of low power factor

Most industrial loads (motors, transformers) are inductive — current lags voltage. This creates a lagging power factor.

Step 2: Add a capacitor in parallel

A capacitor draws current that leads voltage by $90^\circ$. When placed in parallel with the inductive load:

• The capacitor's leading current cancels part of the lagging current
• The net current becomes more in phase with voltage

Step 3: Visualize with phasor diagram

• $I_L$ = load current (lags voltage by angle $\phi$)
• $I_C$ = capacitor current (leads voltage by $90^\circ$)
• $I_{\text{net}}$ = phasor sum of $I_L$ and $I_C$

The resultant $I_{\text{net}}$ has a smaller phase angle $\phi'$ with voltage.

Step 4: Result

The power factor improves (closer to 1). This reduces the current drawn from the source for the same real power, thereby reducing transmission losses.

Key result: A capacitor supplies reactive power locally, reducing the reactive current flowing through transmission lines — improving power factor and efficiency.

Here are the common mistakes students make on this Capacitive Reactance and Power Factor question, along with how to avoid each.

Mistake 1: Confusing "Power Loss" with "Power Used"

The Mistake:

Students often think a low power factor means the load is consuming less power, so it should be good for the transmission line. They fail to distinguish between power consumed by the load ($P_{load}$) and power wasted as heat in the transmission wires ($P_{loss}$).

Why it happens:

They see the formula $P = VI\cos\phi$ and think "low $\cos\phi$ = low $P$ = good." They forget that the transmission line is a separate resistor.

How to Avoid (The Correct Logic):

• Step 1: For a given load, the actual power required is $P = V_{rms} I_{rms} \cos\phi$.
• Step 2: If $\cos\phi$ is low, to deliver the same $P$, the current $I_{rms}$ must be higher (since $P$ is fixed by the load's needs).
• Step 3: Power loss in the transmission line (resistance $R$) is $P_{loss} = I_{rms}^2 R$.
• Conclusion: Higher current $\implies$ much higher $I^2R$ loss. The power factor being low is bad because it forces the line to carry more current than necessary.

Key takeaway: Low $\cos\phi$ $\implies$ High $I$ $\implies$ High $I^2R$ loss.

Mistake 2: Thinking a Capacitor "Cancels" the Load

The Mistake:

Students say the capacitor "cancels" the inductive load, implying the circuit becomes purely resistive and draws zero reactive power. While the net effect is similar, the mechanism is wrong.

Why it happens:

They memorize "capacitor improves power factor" without understanding the phase relationship.

How to Avoid (The Correct Logic):

• Step 1: Most industrial loads (motors, transformers) are inductive — current lags voltage.
• Step 2: A capacitor draws current that leads voltage by $90^\circ$.
• Step 3: The capacitor supplies the lagging reactive power needed by the inductor locally. The source now only supplies real power ($P$).
• Result: The phase angle $\phi$ between total current and voltage decreases, so $\cos\phi$ increases.

Key takeaway: The capacitor doesn't destroy the inductor's reactive power; it provides it locally so the transmission line doesn't have to.

Mistake 3: Forgetting the Capacitor is in Parallel

The Mistake:

Students draw the capacitor in series with the load. This is wrong for power factor correction in transmission.

Why it happens:

They think "capacitor in series" from RC circuits. But series capacitors block DC and change the voltage division — not useful here.

How to Avoid:

• Rule: For power factor correction, the capacitor is always connected in parallel with the load.
• Reason: A parallel capacitor does not change the voltage across the load (it remains the supply voltage). It only changes the total current drawn from the source.
• Exam Tip: Always draw the circuit with the capacitor across the load terminals.

Key takeaway: Parallel connection $\implies$ load voltage unchanged; only source current phase changes.

Mistake 4: Using the Wrong Formula for Capacitance

The Mistake:

Students try to calculate the required capacitance using $X_C = \frac{1}{\omega C}$ but forget to account for the existing reactive power of the load.

Why it happens:

They think "just make $X_C$ equal to $X_L$" — but that's for a series resonance, not power factor correction.

How to Avoid:

• Correct approach: The capacitor must supply the reactive power $Q$ that the load consumes.
• Formula: $Q = V^2 / X_C$ (for parallel capacitor). So, $C = \frac{Q}{\omega V^2}$.
• Where $Q$ comes from: $Q = P \tan\phi$, where $\phi$ is the original phase angle.

Key takeaway: Don't match reactances; match reactive powers.

Quick Summary Table for Revision

Final Exam Tip: When explaining part (b), always mention that the capacitor reduces the phase angle between voltage and current, thereby increasing $\cos\phi$ towards 1.