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As shown the answer is 15 Ohm's . I don't get the trick of the question. can anyone explain to me how we can get the value of R

Thank you all!

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  • \$\begingroup\$ Please read about Wheatstone bridge \$\endgroup\$ – Eugene Sh. Oct 22 '19 at 17:39
  • \$\begingroup\$ Thank you very much, I have not came across this subject before. but it's very clear now this question is a direct application of the Wheatstone bridge. \$\endgroup\$ – user34755 Oct 22 '19 at 17:43
  • \$\begingroup\$ All you need is the bridge equation and solve for the unknown, You should also be able to derive the bridge equation. \$\endgroup\$ – Voltage Spike Oct 22 '19 at 17:57
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You can get the value of R by using the Wheatstone bridge equation.

Wheatstone bridge equation

Which is applicable when the voltage between the two midpoints is zero. So in your case

R = 12/4 * 5 = 15


See more at the wiki page for the Wheatstone bridge: https://en.wikipedia.org/wiki/Wheatstone_bridge

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Well, we have the following circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

The input resistance can be found:

$$\text{R}_\text{in}=\frac{\left(\text{R}_1+\text{R}_2\right)\left(\text{R}_3+\text{R}_4\right)}{\text{R}_1+\text{R}_2+\text{R}_3+\text{R}_4}\tag1$$

Now, the input current is given by:

$$\text{I}_\text{in}=\frac{\text{V}_\text{in}}{\text{R}_\text{in}}=\text{V}_\text{in}\cdot\frac{\text{R}_1+\text{R}_2+\text{R}_3+\text{R}_4}{\left(\text{R}_1+\text{R}_2\right)\left(\text{R}_3+\text{R}_4\right)}\tag2$$

Now, the current \$\text{I}_1\$ is given by:

$$\text{I}_1=\frac{\text{R}_3+\text{R}_4}{\text{R}_1+\text{R}_2+\text{R}_3+\text{R}_4}\cdot\text{I}_\text{in}=\frac{\text{V}_\text{in}}{\text{R}_1+\text{R}_2}\tag3$$

Now, the current \$\text{I}_2\$ is given by:

$$\text{I}_2=\frac{\text{R}_1+\text{R}_2}{\text{R}_1+\text{R}_2+\text{R}_3+\text{R}_4}\cdot\text{I}_\text{in}=\frac{\text{V}_\text{in}}{\text{R}_3+\text{R}_4}\tag4$$

So, we get:

  • $$\text{V}_\text{A}=\text{I}_1\cdot\text{R}_2=\frac{\text{V}_\text{in}\text{R}_2}{\text{R}_1+\text{R}_2}\tag5$$
  • $$\text{V}_\text{B}=\text{I}_2\cdot\text{R}_4=\frac{\text{V}_\text{in}\text{R}_4}{\text{R}_3+\text{R}_4}\tag6$$

So, we also get:

$$\text{V}_\text{A}-\text{V}_\text{B}=\text{V}_\text{in}\cdot\left\{\frac{\text{R}_2}{\text{R}_1+\text{R}_2}-\frac{\text{R}_4}{\text{R}_3+\text{R}_4}\right\}\tag7$$

Using your given values we get:

$$10\cdot\left\{\frac{\text{R}}{5+\text{R}}-\frac{12}{4+12}\right\}=0\space\Longleftrightarrow\space\text{R}=15\space\Omega\tag8$$

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    \$\begingroup\$ +1 for explaining the actual circuit analysis behind the Wheatstone Bridge and not just quoting something from Wikipedia. \$\endgroup\$ – user103380 Oct 22 '19 at 18:42
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    \$\begingroup\$ Far too long elaboration. You can skip all the equations (1) to (4) and start directly with the right hand parts of (5) and (6) and then (7) and (8). \$\endgroup\$ – Huisman Oct 22 '19 at 18:55
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    \$\begingroup\$ You obtain Rin by using a derived rule how to calculate parallel resistors. I think using the derived rule for voltage dividing is way faster to solve this problem. If "good manner" means without derived rules, but by eg using mesh, you shouldn't use equation (1). \$\endgroup\$ – Huisman Oct 22 '19 at 19:02
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    \$\begingroup\$ This is an appreciable answer but i think it's overly generalized. In the question asked we know the voltage between the points between each pair of resistors is 0. Therefore we know beforehand that the ratio of the voltage dividers must be equal, and so the Wheatstone equation applies. No reason to bring current into the equation in order to solve it. \$\endgroup\$ – Shadetheartist Oct 22 '19 at 22:44
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    \$\begingroup\$ I think solving this way will help students to understand the underlying basics instead of blindly substituting known shortcuts. \$\endgroup\$ – Meenie Leis Oct 23 '19 at 14:54

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