At a junction of three wires, two of the wires have currents of +3A and -6A (+ means towards the junction). The third has an 8A fuse in it. Is the fuse likely to burn out?
I think u have to use Kirchoff's Law here, about all currents adding up to 0, but how does that answer the question?
Please explain?
2007-09-04 23:37:03 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Physics
yes, we apply KCL here.
I1+I2+I3=0; I1=3A, I2= -6A;
I3=3A THEN.
no the fuse will not burn out because it can tolerate a current of 8A through it while you are sending 3A current through it.
but if you reverse the direction of either I1 or I2 then we get I3 as 9A(sign irrespective) which will burn the fuse.
2007-09-04 23:47:08 · answer #1 · answered by karan s 3 · 0⤊ 0⤋
Draw a Y shape on a piece of paper. Now put an arrow pointing away from the left arm of the Y and label it -6. On the right arm draw an arrow pointing towards the junction and label it +3. To make everything add up, the lower arm (with the fuse in it) must be +3. The junction is at zero. There is only 3A flowing through the fuse, so the fuse will not burn out.
2007-09-04 23:57:19 · answer #2 · answered by Michael B 6 · 0⤊ 0⤋
Or Norton's theorem or Maxwell's Mesh.
But basically the sum of all the currents must be equal to the current flowing from the supply/battery.
2007-09-04 23:52:38 · answer #3 · answered by Anonymous · 0⤊ 0⤋