Recent content by thisisfudd

So that means that the total capacitance is 7?

So using these formulas I get the right answer for the first resistor. But when I plug back into get I2, I instead get the current for resistor number 3. Any ideas?

So is the current "in the resistor" the same as the current going around the loop?

Hi, ok so if I2 in the top and bottom equations are different, how can I solve for, for instance, I2? Sorry, I should have mentioned that the positive terminals point outward on each battery. So given that, I'm really not sure how I keep messing this up.

If 26 V is applied across the following network, calculate the voltage across each capacitor: Ignore the dots, it's just spaces between the two ends ___ 3.00 uF _____ 4.00 uF |.......| |.......| __________2.00 uF______ |.......| |.......| |.......| So I know that for capacitors in...

OK, I think I get what you're saying. So I did this: 9V + 25I1 + 18I2 = 0 21V + 25I1 + 35I3 = 0 12V + 35I3 - 18I2 = 0 I1 = (-9 - 18I2) / 25 = -.36 - .72I2 I3 = (-12 + 18I1) / 35 = -.34 + .51I2 21 + 25 -.36 - .72I2) + 35 (-.34 + .51I2) = 0 I2 = .67 Does this look better? I...

Ugh, I am totally stressed out by this. Here is what I've done: 9V + 25I1 + 18I2 = 0 12V + 9V +25I1 + 35I3 = 0 12V + 35I3 + 18I2 = 0 I2 = (-9 - 25I1) / 18 = -.5 - 1.39I1 I3 = (-21 - 25I1) / 35 = -.6 - .71I1 12 + 35 (-.6 - .71I1) + 18 (-.5 - 1.39I1) = 0 -21 - 25.56I1 - 9 - 25.02I1 =...

Yeah, I get that. But could you tell me if I am going in the right direction? Because I tried doing that and my answers are far larger than those I expect.

OK, so the current provided by the battery will change: it will decrease. I think the current in the 6 ohm resistor might increase because it is no longer connected in parallel to anything else. Is this correct?

OK I still do not understand AT ALL. I don't understand how I am supposed to add volts and ohms together. I tried setting up two equations whose sums equal zero and then substituting, but it didn't work? 9 V + (I1 x 25 ohms) + (I1 + 18 ohms) = 0 9 V + 12 V + (I1 x 25 ohms) + (I2 + 35 ohms)...

A 3 ohm resistor is connected in parallel with a 6 ohm resistor. This pair is then connected in series with a 4 ohm resistor. These resistors are connected to a battery. What will happen if the 3 ohm resistor breaks? I think that the power dissipated in the circuit will increase. The...

Huh. Awesome. Thanks for your help!

OK, so I have found resistance, .00867. So then I can use P = RI^2? P = .00867 x (15.0A)^2? Your PS is intriguing but of course I don't understand. Are you saying that it draws 15 A on a 120 V line but I have to find what it draws on this line, given the voltage of this line? How would...

An extension cord made of two wires of diameter 0.129 cm (no. 16 copper wire) and of length 2.7 m is connected to an electric heater which draws 15.0 A on a 120V line. How much power is dissipated in the cord? Do I just use P=IV? But that leaves a lot of "extraneous" information. My...

So this problem asks you to determine the magnitudes and irections of the currents in the resistors. I definitely understand how to determine directions, using Kirchhoff's Rules. But here is where I get stuck: the batteries have emfs of E1 = 9.0 V and E2 = 12.0 V and the resistors have vlaues...