PHYSICS

JAMB 2007 - Question 26

Physics 2007 JAMB Past Questions - Question 26: A 40 kW electric cable is used to transmit electricity through a resistor of resistance 2.0 Q at 800 V. The power less as internal energy is

Choose the correct answers from the options given.
A 40 kW electric cable is used to transmit electricity through a resistor of resistance 2.0 Q at 800 V. The power less as internal energy is
A:
B:
C:
D:
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Correct Answer

C

Explanation

To find the power loss as internal energy in the cable, you can use the formula for power dissipation in a resistor:

Power (P) = (Voltage (V))^2 / Resistance (R)

In this case, the voltage (V) is 800 V, and the resistance (R) is 2.0 ohms (2.0 Ω). We can plug these values into the formula:

P = (800 V)^2 / 2.0 Ω

P = (640000 V^2) / 2.0 Ω

Now, calculate the power:

P = 320000 W

The power loss in the cable is 320,000 watts (or 320 kW). This power is lost as internal energy due to the resistance in the cable.

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