Recent content by n387g

Okay, thanks! I just want to double check one more thing. When solving for the currents for each resistor, The current over R1=64 would be I1-I2? And then the current over R=120 and R=82 would be the same, I1? While the current over R=25 and R=110 would be the same, I2?

I'm still having trouble. Based on your last post I formulated these two equations based on the two loops (I went counterclockwise, but that should only affect sign not quantity): For loop 1: 58V-(I1*120)-(I1*82)-[(I1-I2)*64]=0 For loop 2: 3-(I2*25)-[(I2-I1)*64]-(I2*110)=0 I then...

I'm still at a lose of what to exactly do, when approaching this problem mathematically. I went ahead and tried to calculate the current of just R1=64, R2=82, R3=64 and the battery voltage of 58V since if you just consider those three resistors they are in series and therefore share the same...

Homework Statement Calculate the currents in each resistor in this figure: http://session.masteringphysics.com/problemAsset/1084690/2/GIANCOLI.ch26.p32.jpg Homework Equations Kirchhoff's rules In a closed loop: the sum of voltage is 0 A a junction: the sum of current is 0 The...

Using the equation where p=Ql

it's a lowercase L, for length

Okay, so you have to add the electric fields at the +Q, -2Q and +Q. So for +Q, E=k(Ql/(l2+r2)^(1/2)) (cos(theta)rx+sin(theta)ry) And the for -2Q, E=k(2Ql/r2)(sin(theta)ry) And for the other +Q, E=k(Ql/(l2+r2)^(1/2)) (-cos(theta)rx+sin(theta)ry) Is this the correct way to do this?

One type of electric quadrupole consists of two dipoles places end to end with their negative charges (say) overlapping; that is, in the center is -2Q flanked (on a line) by a +Q to either side. Determine to electric field, E, at points along the perpendicular bisector and show that E decreases...