circuit analysisohmsparallelresistanceseries
All resistance are equal in value, \$R1 = R3 = R4 = R5 = R6 = 1~\Omega\$
My calculations are as follows:
I am getting a total resistance equal to \$0.2~\Omega\$, but my teacher said that the correct answer is \$\dfrac{1}{4} = 0.25~\Omega\$. What am I doing wrong?
the book is pretty correct as the circuit is like this:
simulate this circuit – Schematic created using CircuitLab This circuit would be in series if there were no wires shorting the circuit and then net resistance would be 15 ohm.
You are assuming a constant current source and no load on each branch. In that case yes your answer would be right.
However, consider this circuit:
simulate this circuit – Schematic created using CircuitLab
The current through any one resistor (or branch of the circuit) will be \$I=V/R = 10/1000 = 0.01\$
The total current drawn from the power supply will of course be 4x that since there are 4 branches, all at 10mA.
Now, remove one branch, and what happens? Each branch still has \$I=V/R = 10/1000 = 0.01\$ since that hasn't changed at all. What has changed though is the total current. Since there are now only 3 branches the total current drops to 30mA, as that is 3 x 0.01.
The only ways the current through any branch could change is if a) the resistance in that one branch changes, or b) the voltage of the power source changes.
Best Answer
Well, you have the good method but, it seems you have some difficulties to find the equivalent resistor of 2 resistors in parallel.
$$ 1 / Req = 1/R1 + 1/R2 $$ $$ 1 / Req = (R2 + R1) / R2R1 $$ $$Req = R2R1/(R1 + R2)$$
If R1 = R2 = 1 ohm, Req is 0,5 ohm. I let you finish your exercice ;)
simulate this circuit – Schematic created using CircuitLab
Isn't it clearer?