How Can You Calculate the Total Equivalent Resistance in a Circuit?

[HEYJOHN]

What single, equivalent resistor could replace all of the resistors in this circuit?

Rtot = Ω
2 NO

HELP: If you study the diagram, you will see one pair of resistors which is in parallel. Replace this pair with a single equivalent resistor and re-draw the circuit.

HELP: If you have followed the previous suggestion, you will now see a pair of resistors which is in series. Replace this pair with a single equivalent resistor and re-draw the circuit again.

here is an image of the circuit: https://tycho-s.physics.wisc.edu/cgi/courses/shell/common/showme.pl?courses/phys104/fall09/homework/04/r2/ex1s95p3.gif

Req parallel: 1/R1 + 1/R2 = 1/Req

Req series: R1 + R2 =Req

R1=2 R2=5 R3=13 R4=8

I've tried to redraw the circuit where Req= R1 + R2 + [1/(1/R3)+(1/R4)] but that didn't work, any suggestions to which two resistors in the image are parallel to one another to get me started on redrawing the circuit to solve the problem?

 

[HEYJOHN]

Just figured it out, thanks!

 

[tomcue]

Upon studying the circuit, it appears that R3 and R4 are in parallel with each other. Therefore, we can replace them with a single equivalent resistor, let's call it R34. This would result in a circuit with two resistors in series, R1 and R2, and one resistor in parallel with them, R34. We can then use the equations for equivalent resistors in series and parallel to calculate the single equivalent resistor for the entire circuit.

R34 parallel: 1/R3 + 1/R4 = 1/R34

R34 series: R34 + R34 = Req

Solving for R34, we get: R34 = (R3*R4)/(R3+R4) = (13*8)/(13+8) = 5.2 Ω

Now, we can redraw the circuit with R34 replacing R3 and R4:



We can then use the equation for equivalent resistors in series to calculate the single equivalent resistor for the entire circuit:

Req = R1 + R2 + R34 = 2+5+5.2 = 12.2 Ω

Therefore, the single equivalent resistor that could replace all of the resistors in this circuit is 12.2 Ω.