Respuesta :
Okay so here's the approach I took:
The potential difference in each of the circuits must be the same so if we derive equations for both the potential differences we can set them equal to each other and solve for R1:
In the first circuit
V = 2.2(R1)
In the second we have to find the equivalent resistor, since they are connected in series:
1/R1 + 1/R2 + 1/R3... = Rt
We have R2 so...
1/R1 + 1/3.1 = Rt
1/R1 + 0.323 = Rt
So...
V = 1.4(1/R1 + 0.323)
Set those equal:
2.2R1 = 1.4(1/R1 + 0.323)
2.2R1 = 1.4(1/R1) + 0.4522
Now multiply everything by R1 so we can combine like terms:
2.2R1^2 = 1.4 + 0.4522R1
Isolate to form a quadratic
2.2R1^2 - 0.4522R1 - 1.4 = 0
Solving this quadratic:
R1 = 0.90708 or R1 = -0.701
Since R cannot be negative
R1 = 0.907 ohms
Resistance is a type of opposition force due to which the flow of current is reduced in the material or wire. The value of R₁ will be 0.907 ohms.
What is resistance?
Resistance is a type of opposition force due to which the flow of current is reduced in the material or wire. Resistance is the enemy of the flow of current.
The given data in the problem is;
For the first circuit;
V=2.2(R₁)
If the resistors are connected in the series then the value of the equivalent resistance is;
[tex]\frac{1}{Req} =\frac{1}{R_1} +\frac{1}{R_2} +\frac{1}{R_3} \\\\ \frac{1}{Req} =\frac{1}{R_1} +\frac{1}{3.1}\\\\ \frac{1}{Req} =\frac{1}{R_1} +0.323\\\\[/tex]
So the value of the obtained voltage will be;
[tex]\rm V=1.4(\frac{1}{R_1} +0.323)\\\\ 2.2 R_1=1.4(\frac{1}{R_1} +0.323)\\\\ 2.2R_1=1.4\frac{1}{R_1} +0.4522[/tex]
By multiplying by R₁ in the equation we get;
[tex]\rm 2.2 R_1^2=1.4+0.54522R_1 \\\\ \rm 2.2 R_1^2-0.54522R_1-1.4=0\\\\ R_1=0.907 \\\\ R_1=-0.7010 \\\\ R_1=0.907[/tex]
Hence the value of R₁ will be 0.907 ohms.
To learn more about the resistance refer to the link;
https://brainly.com/question/20708652
