Recent content by AstroJMT42

So I squared R is is the power loss, not the new power? I think this is where my misunderstanding is coming from. I thought P=VI and P=I2R were interchangeable. We were given original power and had to use P=I2R to find what the power is after traveling down the line. divide this output power by...

50 Ω is the resistance in the cable. 2,500,000W originally - 500,000 = 2,000,000W sorry that was a silly arithmetic error. So that makes 500,000 / 2,500,000 x 100 = 20%?

Homework Statement Power station produces AC power at 2.5Mw, I = 100A. Transferred along 100km line with a resistance of 50 ohms to a factory. Calculate power lost and % efficiency Homework Equations P=VI P = I2R The Attempt at a Solution Initial power at the station is 2.5x106 . Using I2R...

I just built the circuit using an online tool and the brightness of 6 indeed does not change, I think I understand why now. Thanks for the hints. It also shows that bulb 4 should actually dim, as its flowing from + to - does this mean the bulb dims because it is no longer getting the extra...

oh! So if the voltage and the resistance are the same, then using V=IR the current must be the same, therefore the brightness doesn't change for bulb 6? I'm still not sure whether or not 4 changes though, it is still in series with bulb and the circuit splits before it gets to 5, so would it...

So with bulb 6: In a parallel circuit the amps are shared and the voltage is equal. with S1 shut the current is now split only twice as apposed to 3 times, as the voltage is the same the bulb should be brighter because there are more amps? Bulb 4 is still perplexing me, most likely because it...

Homework Statement All bulbs are identical, (no information given on resistance so assuming all the same) All the switches are closed and then S1 is opened, Does bulb 6 get brighter, dimmer or stay the same? Does bulb 4 get brighter, dimmer or stay the same? Explain briefly why. Homework...