Vidyalelo
Electrical Engineering · Q15

Energy And Power

Engineering and GATE · Electrical Engineering · question 15

Q15

A 120 Ω resistor must carry a maximum current of 25 mA. Its rating should be at least

A.
4.8 W
B.
150 mW
C.
75 mW
Answer
D.
480 mW

Answer: Option C

Solution

Answer: Option C
Solution:
We need to find the power rating of the resistor.
The formula we'll use is: P = I2R, where:

* P is the power (in Watts)
* I is the current (in Amperes)
* R is the resistance (in Ohms)

First, convert the current from milliamperes (mA) to amperes (A):
25 mA = 25 / 1000 A = 0.025 A

Now, plug the values into the formula:
P = (0.025 A)2 * 120 Ω
P = 0.000625 * 120 W
P = 0.075 W

Convert Watts(W) to milliWatts(mW):
0. 075 W = 0.075 * 1000 mW = 75 mW

So, the resistor rating should be at least 75 mW.