Series Parallel Combination?

One way to do this is to set up loop equations for each mesh. Using the familiar Kirchoff laws we get

10=20i1-10i2-10i3 (1)

0=-10i1+30i2-10i3 (2)

0=-10i1-10i2+30i3 (3)

divide through by 10 to simplify

what we want to do is solve for i1 and the resistance is v/i1

to get i1 perform the following row operations

(1)-(3)

1=3i1-4i3

next (2)+3*(3)

0=-i1+3i2-i3+(-3i1-3i2+9i3)==-4i1+8i3 ==>i3=1/2i1

putting this into the above we get

1=3i1-4(1/2)i1=3i1-2i1=i1

Therefor i1=i and the input resistance is 10v/1 A=10Ω

The problem is that the topology needs to be understood.

I see a 10 Volt source of Direct Current.

Leaving the top of the battery we encounter two resistors in parallel that are each 10 ohms.

Each of the 10 ohm resistors are connect in series with another 10 ohm resistor.

The parallel circuits then recombine and return to the other side of the battery.

There is a 10 ohm resistor that interconnects the two halves of the parallel circuit but since there is no possible voltage differential between the points that it is attached there is not current across the resistor.

Each leg of the parallel circuits is equivalent to a 20 ohm resistor.

The combination of the two legs is equivalent to a 10 ohm resistor.

labeling each resistor with a different name we can then determine the voltage drop across each.

Let A and B be the two resistors at the top.

Let C be the horizontal resistor.

Let D and E be the two resistors at the bottom.

Append an "i" to the letter and have that stand for the current across the resistor.

The battery supplies 10 volts

10*Ai + 10*Di=10 volts

Ai=Di

20Ai=10

Ai=.5 amps

10*Bi + 10*Ei=10 volts

Bi=Ei

20*Bi=10

Bi=.5 amps

Ai + Bi= 1 amp

a 10 ohm resistor would also result in a 1 amp current.

Very true, the middle resistor has no effect on the total resistance. look at it this way: the electrons have no reason for passing through a resistor when there is no need to if they want to complete the circuit.

Normally this problem is a bit involved. However in this case the diagonal products are equal. This is to say that the resistor in the middle has no voltage across it.

This problem will have have the same answer if you disconnect the middle resistor.