An electrical power line, having a total resistance of 2 ohms, delivers 1 kW at 220 V. The efficienc

Q: Why do we use $P = I^2 R$ to find the power loss in the wires instead of $P = V^2 / R$?

This is a very common trap! The voltage $V = 220\text{ V}$ given in the problem is the voltage *across the load*, not the voltage drop across the transmission wires. However, because the wires and the load are in series, the current $I$ flowing through the load is the exact same current flowing through the wires. Therefore, $P = I^2 R$ is the safe and correct formula to use for the wires.

Q: Isn't the input power just $1000\text{ W}$? Why do we add the power loss to it?

Think of the power source as a bank account. The load 'withdraws' $1000\text{ W}$, but the transmission line charges a 'transaction fee' of $41.32\text{ W}$ in the form of heat. For the load to successfully receive its $1000\text{ W}$, the source must output the total amount: $1000 + 41.32 = 1041.32\text{ W}$. Efficiency compares what you actually get (the $1000\text{ W}$) to what you originally had to spend (the $1041.32\text{ W}$).

Q: Is the voltage at the source also $220\text{ V}$?

No, it's slightly higher. The load receives $220\text{ V}$, but there is a small voltage drop across the transmission wires equal to $I \times R$. So the source voltage must be $V_{\text{source}} = 220 + (I \times R)$ to ensure exactly $220\text{ V}$ reaches the load.

Q: What if the problem gave the voltage at the source instead of the load?

If the $220\text{ V}$ was the source voltage, the calculation would be more complex. You would have to set up an equation where the total power generated is $V_{\text{source}} \times I$, and this equals $I^2 R_{\text{line}} + P_{\text{load}}$. You'd end up solving a quadratic equation for $I$.

Q: Why is the efficiency so high in this problem?

The efficiency is high because the resistance of the line ($2\,\Omega$) is relatively small compared to the effective resistance of the load. In real life, to keep efficiency high over hundreds of kilometers, power companies use very high voltages to keep the current $I$ extremely small, minimizing the $I^2 R$ losses.

Question

An electrical power line, having a total resistance of $2\Omega$, delivers $1	ext{ kW}$ at $220	ext{ V}$. The efficiency of the transmission line is approximately

Options:

(A) 91%

(B) 85%

(C) 96%

(D) 72%

Sigma Percentile · Accepted Answer

Imagine you are an engineer tasked with delivering electricity from a power plant to a distant town. The power plant generates the electricity, but the wires carrying it aren't perfect—they have resistance. This means that as the electrical current fights its way through the wires, some of the precious energy is lost as heat. This problem is a classic exploration of that exact scenario, testing your understanding of power, current, and efficiency.

### Analyzing the Setup

We are given a load that requires a specific amount of power to function properly: $P_{	ext{out}} = 1	ext{ kW} = 1000	ext{ W}$. It operates at a voltage of $V = 220	ext{ V}$.

The transmission line connecting the source to this load has a total resistance of $R = 2\,\Omega$. Our goal is to find out how efficient this entire setup is. Efficiency is simply a measure of how much of the total power generated actually makes it to the load to do useful work.

### Finding the Current

Before we can figure out how much power is lost in the wires, we need to know how much current is flowing through them. The current is dictated by the load's requirements.

Using the fundamental power equation $P = VI$, we can rearrange it to solve for current:

$$I = \frac{P}{V}$$

Substituting our known values:

$$I = \frac{1000}{220} = \frac{50}{11}	ext{ A}$$

This is the current that must flow through the entire circuit—from the source, through the transmission lines, and into the load.

### Calculating the Power Loss

Now that we have the current, we can calculate the toll the transmission line takes. As this current $I$ flows through the resistance $R$ of the wires, energy is dissipated as heat. This is known as Joule heating, and the formula is:

$$P_{	ext{loss}} = I^2 R$$

Let's plug in our current and resistance:

$$P_{	ext{loss}} = \left(\frac{50}{11}ight)^2 	imes 2$$

$$P_{	ext{loss}} = \frac{2500}{121} 	imes 2 = \frac{5000}{121} \approx 41.32	ext{ W}$$

So, the wires are acting like a $41.32	ext{ W}$ heater, stealing energy before it reaches the destination.

### Determining the Efficiency

Efficiency $\eta$ is defined as the ratio of the useful output power to the total input power provided by the source.

$$\eta = \frac{P_{	ext{out}}}{P_{	ext{in}}} 	imes 100\%$$

**Here is where many students make a silly mistake.** They assume the input power is just $1000	ext{ W}$. But the source doesn't just power the load; it also has to supply the power that gets lost in the wires! Therefore, the true input power is the sum of the output power and the power loss.

$$P_{	ext{in}} = P_{	ext{out}} + P_{	ext{loss}} = 1000 + 41.32 = 1041.32	ext{ W}$$

Now, we can calculate the final efficiency:

$$\eta = \frac{1000}{1041.32} 	imes 100\% \approx 96.03\%$$

Rounding to the nearest whole number, we get **$96\%$**.

This is a fantastic result, but it also reveals a profound truth about real-world power grids. If we tried to transmit megawatts of power at a low voltage like $220	ext{ V}$, the current would be massive, and the $I^2 R$ losses would melt the wires! This is exactly why power companies use transformers to step up the voltage to hundreds of thousands of volts for long-distance transmission. High voltage means low current, and low current means minimal power loss.

Animated Solution for Physics - Current Electricity: An electrical power line, having a total resistance of $2Ω$ , delivers $1 kW$ at $220 V$ . The efficiency of the transmission line is approximately

Select Answer:

Visualized Solution

Visualizing the Transmission Setup

Formula for Current

Substituting Values for Current

Calculating the Current

Formula for Power Loss

Substituting Values for Power Loss

Calculating Power Loss

Formula for Efficiency

Substituting Values for Efficiency

Calculating Final Efficiency

Final Answer

The Way Forward

The Sigma Insight: Ohm's Law, Resistance and Electrical Power

Analyzing the Setup

Finding the Current

Calculating the Power Loss

Determining the Efficiency

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