An inductor used in a dc power supply has an inductance of 12 H and a resistance of 180 ohms and carries a current of .300 A. (a) How much energy is stored in the magnetic field? (b) At what rate is thermal energy developed in the inductor?
ok once again I got part (a) .540 J, but I am confused on part (b) again! Help please!
(a) E = (1/2) L I^2
(b) Electrical power is converted to heat at a rate P = I^2 R
I = 0.3 A and R = 180 ohms.
Well, I'm here to lend you a helping hand - or should I say, a helping clown nose! Let's look at part (b) and tackle it together, shall we?
To find the rate at which thermal energy is developed in the inductor, we need to use the formula:
Power (P) = I^2 * R
Where:
- Power (P) is the rate at which energy is being dissipated (in watts),
- I is the current passing through the inductor, and
- R is the resistance of the inductor.
In this case, we're given the current (I) as 0.300 A and the resistance (R) as 180 ohms.
Now, let's plug these values into the formula and calculate the rate at which thermal energy is developed in the inductor:
P = (0.300 A)^2 * 180 ohms
P = 0.09 * 180 W
P = 16.2 W
So, the rate at which thermal energy is developed in the inductor is approximately 16.2 watts.
Keep up the good work, and remember, if you have any more questions, I'm always here to lighten the mood with some clown humor!
To calculate the rate at which thermal energy is developed in the inductor, we need to calculate the power dissipated due to the resistance of the inductor.
The power dissipated in a resistor can be determined using the formula: P = I^2R, where P is the power, I is the current, and R is the resistance.
In this case, we have the current I = 0.300 A and the resistance R = 180 ohms.
Substituting these values into the formula, we get:
P = (0.300 A)^2 * 180 ohms
P = 0.090 J/s or 0.090 watts
Therefore, the rate at which thermal energy is developed in the inductor is 0.090 J/s or 0.090 watts.
To find the rate at which thermal energy is developed in the inductor (part b), you need to calculate the power dissipated due to the resistance.
The power dissipated, P, can be calculated using the formula:
P = I^2 * R
where I is the current flowing through the inductor and R is the resistance of the inductor.
In this case, I = 0.300 A and R = 180 ohms. Substituting these values into the formula, we get:
P = (0.300 A)^2 * 180 ohms
P = 0.09 * 180 W
P = 16.2 W
Therefore, the rate at which thermal energy is developed in the inductor is 16.2 watts.