An electric generator supplies power to a workshop through transmission lines having a tot… — JAMB UTME

An electric generator supplies power to a workshop through transmission lines having a total resistance of $2\,\Omega$ . The workshop operates electrical equipment drawing a power of $4\,\text{kW}$ at a terminal voltage of $200\,\text{V}$ . What is the total power generated by the generator?

$4.8\,\text{kW}$ Cevap
B
$4.0\,\text{kW}$
C
$4.04\,\text{kW}$
D
$24.0\,\text{kW}$

Cevap

The total power generated by the generator is

4.8\,\text{kW}

The electrical current flowing through the system is

20\,\text{A}

based on the load power of

4000\,\text{W}

200\,\text{V}

. The power lost in transmission lines is

P = I^2 R = 20^2 \times 2 = 800\,\text{W}

(

0.8\,\text{kW}

). Therefore, the total electrical power generated by the source must be the sum of the load power and line losses, giving

4.0\,\text{kW} + 0.8\,\text{kW} = 4.8\,\text{kW}

Adım Adım Çözüm

Calculate the current flowing through the circuit using the power and voltage at the workshop.

I = \frac{P_{\text{workshop}}}{V} = \frac{4000\,\text{W}}{200\,\text{V}} = 20\,\text{A}

Current is uniform across the series transmission line.

Calculate the power loss dissipated as heat in the transmission lines.

P_{\text{loss}} = I^2 R = (20\,\text{A})^2 \times 2\,\Omega = 400 \times 2 = 800\,\text{W} = 0.8\,\text{kW}

By Joule's law of heating, power dissipated in a resistor carrying current

I

I^2 R

Sum the useful power delivered to the workshop and the transmission power loss to obtain total generated power.

P_{\text{total}} = P_{\text{workshop}} + P_{\text{loss}} = 4.0\,\text{kW} + 0.8\,\text{kW} = 4.8\,\text{kW}

Total energy generated per unit time equals energy consumed by load plus energy lost.

Anahtar Kavram

Power Loss in Transmission Lines and Total Source Power

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