If you are given two pairs of parallel circuits, with three unknown resistors and one 4ohm resistor, with the total resistance totaling 3/4 ohms ; how are you supposed to find the resistance the unknown resistors? Do you just use the parallel resistor formula : rt = 1/r1 + 1/r2 + 1/r3 ... But using that formula will simply give you the total of the three unknown resistor's resistance, not the resistance of each resistors. How do you go about to solve for the unknown resistors. Note that the only numbers given are one of the resistor and the total resistance. Thanks, There's a picture below of the diagram


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ that is one pair of parallel circuits .... one parallel circuit on top and one on bottom \$\endgroup\$ – jsotola Apr 28 at 1:42
  • \$\begingroup\$ Looks like all three unknown resistors are supposed to be the same value. Is that the case here? \$\endgroup\$ – jonk Apr 28 at 2:18
  • 1
    \$\begingroup\$ You solve it via the following quadratic:$$\begin{align*}R_\text{TOTAL}&=\frac{1}{\frac{1}{R_1}+\frac{1}{R=4\:\Omega}}+\frac{1}{\frac{1}{R_1}+\frac{1}{R_1}}=\frac34\:\Omega\\\\&=\frac{R\,R_1}{R+R_1}+\frac{R_1}{2}\\\\&=\frac{R_1^{\,2}+3\,R\,R_1}{2\left(R+R_1\right)}\\\\&\therefore\\\\0&=R_1^{\,2}+\left(3\,R\,R_\text{TOTAL}-2\,R_\text{TOTAL}\right)R_1-2\,R\,R_\text{TOTAL}\end{align*}$$You know \$R=4\:\Omega\$ and \$R_\text{TOTAL}=\frac34\:\Omega\$. You should be able to solve this for \$R_1\$. \$\endgroup\$ – jonk Apr 28 at 2:48

Its easy you 2 parallel circuit attached in series.

First divided them into tow blocks First block is = ( R1*4ohms)/(R1+4ohms) second block is = (R1*R1)/(R1+R1)

after getting the result of the first block and the second block add them together it becomes something like this Rt=((R1*4)/(R1+4)) + ((R1*R1)/((R1+R1)) R1 have the same value. Let's assume that you have the voltage and current.

V=RT*I ............ RT=V/I ........... RT-4 = new RT ... ............ R1=new RT/3. = value of each r1


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