Recent content by Numbskull

Thank you Babadag. I'm on to the next part of the question now which asks what value resistor would need to be placed in the field circuit (in addition to the 220Ω) to increase the motor speed to 1000rpm with the same load torque. I know that reducing the current in the field winding, and thus...

Unfortunately I am not provided with a number of motor parameters so am unable to use those formulas.

Yes, I've got the answer with little effort to be honest. As far as I can see, the need to mention 20kW in the question appears immaterial, which is what confused me.

I surprisingly got a rapid email reply from my tutor who confirmed that in this case, it was the armature current half way between no-load and full load.

Unfortunately the notes only provide a single example which applies to a generator. No mention of half-load or motors.

Was just checking that I'm on the right lines. So (approximate figures): Field winding loss is fixed at 1136 Watts, where the current is 2.27 Amps. Motor current at no-load is (5 - 2.27) = 2.73 Amps. Therefore, excluding the field winding current, we have 42.72 - 2.72 = 40 Amps in the...

First some basic figures which are very rounded as I'm interested in the approach to the problem rather than accuracy of the answer at this stage: The field current will be 500/220 = 2.2727 Amps, and so power is a fixed loss at 1136.36 Watts The armature current is 45 - 2.2727 = 42.72 Amps. At...

I have done this question and can tell you that I couldn't get the thing to work at all with some zero values. As scottdave above says, you need to put in a very small number just to allow it to compute. Having just looked at my tutor feedback document, I did get the phases incorrect in my...

Thank you rude man. Following your reasoning, I have a total secondary impedance of (##605\Omega##), minus the impedance of the load (##592.9\Omega##), minus the reflected primary impedance (##7.0257\Omega##). As the question asked for the maximum value of the secondary winding resistance, I...

So I have ##R_p = \sqrt {0.01^2 + 0.057^2}## which is ##0.05787 \Omega## Multiplied by ##n^2 (702.569)## is ##40.658\Omega## With a secondary output of 11kV and a 200kVA rating, the (max) current is 18.1818 Amps. With a voltage regulation of 2%, I have 11kV * 0.98 which is 10.78kV (across the...

Yes, the correct final position of the slider is 0.5. I then fiddled with the equation I posted to get ##-50x^2 + 125x - 50 = 0## and popped that into the quadratic formula for which there is only one credible answer.

I did this some time ago, and I feel your pain with the learning materials, they are woefully inadequate! I called the parallel resistors ##R_o## and followed with ##R_o = \frac{10 \left(1 - x\right) \text{ x } 5}{10 \left(1 - x\right) + 5}## Does that help?

Was trying to follow, but got a bit lost. I get that ##\frac{Is}{Ip}=\frac{N_1}{N_2}##, but couldn't see how ##Ps = \left[\frac{N_1}{N_2}\right]^2## I'm sure it is just an algebra manipulation, but with some intermediate steps left out.

So far then I have primary impedance: ##Z = \sqrt{0.01^2 + 0.057^2}## which equals ##0.05787\Omega## As ##R's = Rs + \frac{Rp}{n^2}## then ##R's = Rs + \frac{0.05787}{702.569}## is ##R's = Rs + 82.369\mu\Omega## Am I on the right track? That primary reflected result (##82.369\mu\Omega##)...

And simply using ##\frac{N1}{N2}## or ##\frac{N2}{N1}## is correct for step-up and step-down respectively?