Recent content by Daltohn

Yep, thank you all :)

Okay, I'm an idiot... Suppose s<t. Then t is not in the same row as s since none of its members is greater than s. But then t can never be the least in its column since every column has a member that is less than or equal to s. However, t is by definition the least in its column. Contradiction...

I don't know except that they're all less than s. What more can I deduce assuming s<t?

Homework Statement The numbers 1 to 25 are arranged in a square array of five rows and five columns in an arbitrary way. The greatest number in each row is determined, and then the least number of these five is taken; call that number s. Next, the least number in each column is determined, and...

Ok haha, good! Thanks!

I'm just lost at this point. How can current go downward through the inductor when the circuit is in steady conditions, does conventional current not go from positive to negative? The current then goes upwards through the resistor as well, and when the switch opens the current's direction...

The upper end of the resistor is positive since current flows downward. Then it would make more sense to me that the upper end of the inductor is positive because the emf increases the potential. Can I compare it to the motional emf of a moving rod? In the picture, current flows...

Thanks for helping to clear this up! :) In the original post it says the answer is 2.3 kV but it should be 1.3 kV.

The formula gives a positive value for the emf which makes sense. But does the upper part of the coil have higher potential?

The induced current should try to counter-act the decrease, so the induced current has the same direction. The emf would then have "the same" direction as the initial emf from the battery. Does the upper part of the inductor have positive potential then? Then the current goes through the...

Self-inductance electromagnet, circuit So when the switch is opened, you shouldn't apply KVL so that emf-125i-4000i=0? Then it would work like a battery with internal resistance. Or is that voltage drop not taken into account here?

(4000+125) of course! I actually thought of that after I did the above, but I didn't reflect over that they give the same value after rounding off, so assumed the first one was right haha. Thanks!

Self-inductance electromagnet, circuit KVL: 40V-12.5I=0 I=0.32 A. I got that far. But now I realized what to do. :) Very soon after K is opened, I is still 0.32 A through the inductor, which now forms a closed circuit with the resistor. The voltage drop over the resistor is V=IR=1.3 kV, which...

Homework Statement The wire in an electromagnet has the resistance 125 ohms and a parallel resistor has the resistance 4 kiloohms. The magnet is kept magnetized by a DC voltage 40V. What is the voltage over the electromagnet soon after switch K is opened? Also comment on the polarity of the...

Aarghh, this book has so many errors... Thanks. :) While we're at it, our equation for work is charge*voltage, so here the voltage would be positive and the charge negative. Is that how you do it? I'm often confused whether work should be subtracted or added but have negative value. With...