Find equivalent resistance of a resistor circuit [closed]

Question

what is the process for finding the equivalent resistance of the following circuit?

Answer 1

By carefully observing the circuit and then rotating it ninety degrees to the right and moving the resistors without modifying the topology of the network, we find that the circuit is a Wheatstone bridge, well known to everyone:

Answer 2

Hint:

Replace the voltage source with a DC 1 Ampere current source with its positive direction upwards as in the figure. Connect the bottom nodes to ground. Assign voltage voltages to three nodes on top such as V1, V2 and V3. Perform nodal analysis and you will get three linear equations. Use Kramer's rule or Gauss jordan method to get the voltages V1,V2 and V3. The voltage V1 on the positive terminal of 1A current source would give you equivalent resistance of the circuit. \$R_{eq}= \frac {V_{1}}{1}\$.

Answer 3

There isn't just one process. There's many different approaches. Some more general than others.

But in this case I'd like to call your attention to the voltage dividers present in your circuit.

Note that \$R_3\$ and \$R_5\$ form one voltage divider and that \$R_1\$ and \$R_4\$ form another voltage divider. So Thevenize \$R_3\$ and \$R_5\$ and then Thevenize \$R_1\$ and \$R_4\$. You now have:

^{simulate this circuit – Schematic created using CircuitLab}

This tells you the two node voltages.

From this, you only need to sum up the two currents, the two present in \$R_1\$ and \$R_3\$ since those are the only current paths that the voltage source can take, and divide that sum into the source voltage, \$V=20\:\text{V}\$:

$$\begin{align*}R_\text{EQ}&=\frac{V}{\frac{V-\left(V_{\text{TH}_1}-I_\text{TH}\,\cdot\,R_{\text{TH}_1}\right)}{R_3}+\frac{V-\left(V_{\text{TH}_2}+I_\text{TH}\,\cdot\,R_{\text{TH}_2}\right)}{R_1}}\\\\&=\frac{20\:\text{V}}{\frac{20\:\text{V}-12\:\text{V}}{10\:\Omega}+\frac{20\:\text{V}-8\:\text{V}}{30\:\Omega}}=\frac{20\:\text{V}}{800\:\text{mA}+400\:\text{mA}}\end{align*}$$

And you are done.