Question 724: Current Electricity

Question

The maximum power rating of a $20 \Omega$ resistor is 2 kW . Then, it cannot be used with 300 V DC source because the

Accepted Answer

The maximum power rating of a $20 \Omega$ resistor is 2 kW. This resistor should not be used with a 300 V DC source because it would exceed its power rating, potentially damaging the resistor due to excessive heat.
Explanation:
Maximum Power Rating
: The resistor's maximum power rating is 2 kW.
Calculating Power Dissipation
:
When connected to a 300 V DC source, the power dissipated across the resistor is calculated using the formula:
$    P = \frac{V^2}{R} = \frac{300^2}{20}    $
Calculation
:
$    P = \frac{300 	imes 300}{20} = 4500 \, 	ext{W}    $
$    P = 4.5 \, 	ext{kW}    $
Comparison
:
Since 4.5 kW > 2 kW, the power dissipation would exceed the resistor's maximum power rating.
Therefore, the resistor is at risk of being damaged because it would produce excessive heat beyond its tolerance.

Answer

current drawn is very large and it can blow the fuse.

Answer

current drawn is very less and appliance does not work.

Answer

resistor may melt due to excessive heat.

Answer

None of the above

Current Electricity

The maximum power rating of a 20 \Omega resistor is 2 kW . Then, it cannot be used with 300 V DC source because the

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