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Let say you have a network of resistors. All resistors have values but one of them is labeled "R"

Given the value of a total voltage and a total current, find the value of R.The total current is 5A and the total voltage is 250V.

My head start would be using Ohm's Law to compute for total resistance. But after that, I don't know what to do. I did tried doing the usual method of solving resistors in series and parallel. But I ended up having a negative answer.

How can I get the value of R?

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    \$\begingroup\$ can you share your calculations? \$\endgroup\$
    – User323693
    Jan 22 '18 at 14:20
  • \$\begingroup\$ Hint: the current \$I_T\$ is the same if 7 ohm and 8 ohm resistors are put directly in series with each other to form a 15 ohm resistor. \$\endgroup\$
    – Andy aka
    Jan 22 '18 at 14:33
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    \$\begingroup\$ Just because you get a negative answer does not necessarily mean it's wrong. For example, what resistor must you put in parallel with a 5K resistor to make it 10K? The answer is -10K, which is not possible with a passive component, but that's the correct answer. \$\endgroup\$ Jan 22 '18 at 14:36
  • \$\begingroup\$ Umar; sorry I can't share calculations. :( I did solved that but ended up not getting the right one. \$\endgroup\$ Jan 23 '18 at 6:05
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Well, starting backwards:

  1. $$\text{R}_1=10+15+\text{R}\tag1$$
  2. $$\text{R}_2=\frac{30\cdot\text{R}_1}{30+\text{R}_1}\tag2$$
  3. $$\text{R}_3=32+25+\text{R}_2\tag3$$
  4. $$\text{R}_4=\frac{60\cdot\text{R}_3}{60+\text{R}_3}\tag4$$
  5. $$\text{R}_5=\text{R}_{\space\text{in}}=8+7+\text{R}_4\tag5$$

So, we get:

$$\text{V}_{\space\text{in}}=250=\text{I}_{\space\text{in}}\cdot\text{R}_{\space\text{in}}=$$ $$\text{I}_{\space\text{in}}\cdot\left\{8+7+\frac{60\cdot\left(32+25+\frac{30\cdot\left(10+15+\text{R}\right)}{30+10+15+\text{R}}\right)}{60+32+25+\frac{30\cdot\left(10+15+\text{R}\right)}{30+10+15+\text{R}}}\right\}=$$ $$225\cdot\text{I}_{\space\text{in}}\cdot\frac{505+11\cdot\text{R}}{2395+49\cdot\text{R}}\tag6$$

So, for example when \$\text{I}_{\space\text{in}}=\text{I}_{\space\text{T}}=5\space\text{A}\$, we get:

$$250=225\cdot1\cdot\frac{505+11\cdot\text{R}}{2395+49\cdot\text{R}}\space\Longleftrightarrow\space\text{R}=245\tag7$$

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The approach I would take to this problem is to work from left to right.

  • We know the total current so we can work out the voltage across the 8 ohm and 9 ohm resistors.
  • We subtract those voltages from the supply voltage to work out the voltage across the 60 ohm resistor.
  • We then work out the current in the 60 ohm resistor.
  • We subtract the current in the 60 ohm resistor from the total current to get the current that flows onward in the circuit.

Repeating this process we can work out all the voltages and currents in the circuit until we have the voltage and current at R. We can then attempt to calculate a resistance.

Note that only a limited range of current values are achievable. If the current specified is too high or too low the calculations will break down (and if you plow ahead anyway you will likely come out with a negative value for R).

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  • \$\begingroup\$ What if I want to redraw the whole diagram? I need to remove the voltage source right? \$\endgroup\$ Jan 23 '18 at 6:06

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