I have the question "For the circuit shown find the value of the current I"
From the previous question before this I worked out V = 60,
Therefore V = IR
And so I = V/R = 60/10 = 6A.
Is this correct ?
Electrical – Finding the value of current I
circuit analysis
Related Solutions
It's quite easy if you exploit the symmetry:
Because of the radial symmetry the 5 marked nodes have the same potential. Therefore there will be no current going throught the 5 marked resistors, i.e. you can treat them as being absent.
So the final resistance will be (R + R / 2) / 5
I'm not going to do your homework for you, but maybe give you different ways of looking at this problem.
Note that this problem really comes down to is finding the apparent resistance across one of the resistors in the network.
These problems are often deliberately presented in a confusing geometry. The first thing you should do is write it down in a more accessible form:
All resistors are equal, which we simply call R. The real question comes down to what is the apparent resistance to a connection across R18. Hopefully you can see that this is a purely passive linear network, so the answer must be xR. Your job is to find x.
This is not a network that can be trivially simplified by inspection. There are various ways to attack it. One method is to solve the resistance of little pieces at a time, then combine them. Another is a numerical relaxation algorithm. It's not going to be simple. Symmetry will help somewhat, but not as much as you seem to expect.
Get crack'n.
Best Answer
No, it's wrong.
Since the current will divide in parallel connections, there are different currents flowing through the branches those are \$I\$1, \$I\$2 and \$I\$3.
\$I\$ is the whole current supplied by the source. So you need to calculate all the branch currents and then add them.
That means you need to use KCL, \$I\$ = \$I\$1 + \$I\$2 + \$I\$3.