A motor attached to a 120 V/60 Hz power line draws an 8.0 A curre...

Question

A motor attached to a 120 V/60 Hz power line draws an 8.0 A current. Its average energy dissipation is 800 W.d. How much series capacitance needs to be added to increase the power factor to 1.0?

Accepted Answer

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let's read the problem and highlight all the key pieces of information that we need to use. In order to solve this problem, an induction motor operates using 230 volts per 50 Hertz electricity. The motor takes in a 12 amps of electric current and expends electrical energy at a rate of 1000 watts. If the power factor of the circuit is to be corrected to 1.0 what capacitor should be used. So that's our angles. We're trying to figure out what the capacitor value should be in order to have a power factor circuit to be corrected to 1.0. So that's our end goal. So we're given some multiple choice answers. Let's read them off to see what our final answer might be. A is 0.46 milli Ferras B is 0.23 Miller C is 89 micro Ferras and D is 0.18 Miller. Awesome. So first off, let us assume that the motor is in a series RLC circuit. Now let us recall and use the phase angle equation that states that cosine of phi is equal to the power source, which we're gonna denote it as P subscript S divided by the R MS current multiplied by the RMS EMF value. So which is represented by epsilon subscript RMS? Awesome. So now we need to isolate and solve for phi is phi by itself is the phase angle. So when we do that, we will get that Phi is equal to the inverse cosine of the power source divided by the R MS current multiplied by the RMS EMF value. Awesome. So this is equal to, when we plug in our known variables to solve for the phase angle, we get that the inverse cosine of 1000 watts which is the power source divided by 12 amps multiplied by the epsilon RMS value which is 230 volts. So when we plug it into a calculator, we'll find that our phase angle is equal to 68.76 degrees, we can also write the phase angle as. So this is an important thing to note, we can also write the phase angle as equal to inverse tangent of the inductive reactant minus the capacitive reactance all divided by R which represents the resistance. Therefore, we can go on to write that the inductive reactance minus the capacitive reactance is equal to R multiplied by the tangent of Phi. So by using Ohm's law, we could solve for the resistance using the following equation that states that are is equal to VR MS which is the R MS voltage divided by the R MS current. So the equation to solve for the resistance voltage or the R MS voltage is given by this is important the R MS. So the R MS voltage is equal to the epsilon R MS value or the ems, the R MS EMF value multiplied by a cosine of phi which when we plug in are known variables will get 230 volts multiplied by cosine of 68.76 degrees. Which when we plug that into a calculator, we will get that VR MS is equal to 83.32 volts. So at this stage, we can solve for the resistance. And let's note that the resistance capital R is equal to VR MS divided by Ir MS which is equal to 80 3.32 volts divided by 12 amps, which is equal to 6.943 ohms. So now we can solve for the inductive reactants minus the capacitive reactants. So let's do that. So when we do that, we get that the inductive reactant subtracted by the capacitive reactants is equal to the resistance multiplied by the tangent of Phi which Phi to reiterate is the phase angle. Awesome. So when we plug in our known variables, we will get 6.943 ohms multiplied by the tangent of 68.76 degrees, which is equal to 17.86 ohms. So note that the phase angle must be zero as the power factor is corrected to 1.0. This is done by increasing the capacitive reactance by 17.86 ohms. So therefore, the capacitance having this reactance will be, and we can write it as the capacitance C is equal to one divided by two multiplied by pi multiplied by 50 Hertz multiplied by 17.86 ohms. So when we plug that into a calculator, we will get 1.78 multiplied by 10 to the power of negative four Ferras. And let's note that will be convert that to milli ferrets. And we round up, we will get 0.18 milli fair ads. And that is our final answer. So that means that the capacitor that should be used will have to have a value of 0.18 Miller adds and we have to. So when we solve for our capacitance, it's given to us in fair ads. So we will have to convert fads to miller ads. And I've gone ahead and speeded up, speeded up ahead to solve for that. But you can use dimensional analysis or you can quickly look up a conversion chart. Awesome. So looking at our multiple choice answers, our correct answer has to be the letter D 0.18 milli Ferras. Thank you so much for watching. Hopefully that helped and I can't wait to see the next video. Bye.

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A motor attached to a 120 V/60 Hz power line draws an 8.0 A current. Its average energy dissipation is 800 W.d. How much series capacitance needs to be added to increase the power factor to 1.0?

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