Electric Circuits
Beginner
Ohm's Law
Source: High school physics (Chinese)
Problem
The total resistance of a transmission line is $1.0$ $\Omega$, and the transmitted power is $100$ kW.
- If the power is transmitted at the low voltage $400$ V, what is the heating loss on the transmission line?
- If the power is instead transmitted at the high voltage $10$ kV, what is the heating loss?
(1) $P_{\text{loss}} = 62.5$ kW (62.5% of transmitted power); \quad (2) $P_{\text{loss}} = 100$ W (0.1% of transmitted power).
For a transmission line carrying power $P$ at line voltage $U$, the line current is $I = P/U$ and the heating loss is $P_{\text{loss}} = I^2 R = (P/U)^2 R$.
This shows that loss scales as $1/U^2$ — raising the voltage drastically reduces transmission loss.
(1) At $U_1 = 400$ V:
$$I_1 = \frac{P}{U_1} = \frac{1.0 \times 10^{5}}{400} = 250 \text{ A}$$ $$P_{\text{loss},1} = I_1^2 R = 250^2 \times 1.0 = 6.25 \times 10^{4} \text{ W} = 62.5 \text{ kW}$$(2) At $U_2 = 10$ kV $= 10^{4}$ V:
$$I_2 = \frac{P}{U_2} = \frac{1.0 \times 10^{5}}{10^{4}} = 10 \text{ A}$$ $$P_{\text{loss},2} = I_2^2 R = 10^2 \times 1.0 = 100 \text{ W}$$Increasing the transmission voltage by a factor of $25$ reduces the loss by a factor of $625$.