Recent content by johndough999

Thanks for the help!

The same way as for the last part, where I find the current in the center branch?

Why can i not just simply use ohms law to find the current for part 2 and 4?

Does that mean the answers are incorrect?

iii) 80 0hms iv) I=V/R I=10/80 I=1/8 A v) 0.02A? since current in the centre branch when V1 is short circuited is 0A.

So if V1 is short circuited and the voltage across it zero, voltage across 60 and 40 ohms is zero therefore voltage across 80 ohm resistor is equal to V2?

the voltage across 40 and 60 ohm would be the supply voltage V2 - voltage across 80 ohm resistor

donpacino could you elaborate a little more. Would the voltage across the 60 and 40 ohm be the total voltage - voltage across 80 ohm?.

Zseen = 80Ω + ((60Ω+40Ω) || ZV1) From the formula you mentioned does that mean 80Ω is in parallel with 60 and 40 Ω?

For part iii) do I not have to short circuit V1? hence would i not be left with a series circuit?

Homework Statement Homework Equations Law of Superposition I=V/R The Attempt at a Solution I am reducing the circuit. i) By law of Superposition, short circuit V2 100 ohms on the leftmost branch and 50 ohms on the rightmost branch, both are in parallel so 100*80/(100+80)=44.4 ohms seen by...

Thanks for the help hesch!

Hi Hesch, to find the distance from the origin are you using the following formula? √ X^2 + Y^2 + Z^2

Homework Statement There exists a current of 5A in the ay direction at x=2m,z=-2m where the ay is the unit vector in the y direction. Find the magnetic field strength at the origin. Homework Equations B = (μI/2πR) * r B = μH The Attempt at a Solution B = (μI/2πr) B= (4π*10^-7)(5)/2π(2) where...