I am preparing for an electrical engineering exam coming up and I got stuck on task C, where you have to determine the resistance between 1 and 2. How do you do that exactly? I have already solved tasks a and b.
Well, I know you have to switch off all sources first and then determine the resistance, but somehow I can't see how I should redraw the circuit to see if it is a series or parallel circuit. I would be grateful for any tips.
Information about the circuit R=10 ohm Uq=20V, Iq= 3A.
Another way to solve this is by solving the nodes' current equations.
First redraw the circuit, number the nodes and their voltages, then write the nodes' equations and solve \$R_{12} = x R = 10x \Omega\$, the resistance asked for.
simulate this circuit – Schematic created using CircuitLab
The sum of currents going into every node is zero.
$$ \frac{U_1}{xR} + \frac{U_3-U_1}{R} + \frac{U_4-U_1}{3R} = 0 \\ \frac{U_3}{3R} + \frac{U4}{R} + \frac{-U_1}{xR} = 0 \\ \frac{U_1-U_3}{R} +\frac{U_4-U_3}{2R} + \frac{-U_3}{3R} = 0 \\ \frac{U_1-U_4}{3R} + \frac{U_3-U_4}{2R} + \frac{-U4}{R} = 0 $$
Recognize that \$U_1\$ is the driving voltage and therefore has to end up at the right hand side of the equations.
$$ \begin{align} \frac{U_3}{R} + \frac{U_4}{3R} &= -\frac{U_1}{xR} + \frac{U_1}{R} + \frac{U_1}{3R} \\ \frac{U_3}{3R} + \frac{U_4}{R} &= \frac{U_1}{xR} \\ -\frac{U_3}{R} -\frac{U_3}{2R} - \frac{U_3}{3R} +\frac{U_4}{2R} &= -\frac{U_1}{R} \\ \frac{U_3}{2R} -\frac{U_4}{3R} -\frac{U_4}{2R} -\frac{U_4}{R} &= - \frac{U_1}{3R} \end{align} $$
Multiply the top \$2\$ equations with \$3xR\$ and the bottom \$2\$ with \$6R\$ to get
$$ \begin{align} 3 x U_3 + x U_4 &= ( -3 + 3x + x ) U_1 \\ x U_3 + 3 x U_4 &= 3 U_1 \\ ( - 6 - 3 - 2 ) U_3 + 3 U_4 &= - 6 U_1 \\ 3 U_3 + ( - 2 - 3 - 6 ) U_4 &= - 2 U_1 \end{align} $$
and after regrouping
$$ \begin{align} 3 x U_3 + x U_4 &= ( 4 x - 3 ) U_1 \\ x U_3 + 3 x U_4 &= 3 U_1 \\ 11 U_3 - 3 U_4 &= 6 U_1 \\ 3 U_3 - 11 U_4 &= - 2 U_1 \end{align} $$
The top 2 equations give \$ -8xU_4 = (4x - 12) U_1 \$, or $$ 2 U_4 = ( 3/x - 1 ) U_1 $$ and the lower 2 equations give \$ 112 U_4 = 40 U_1 \$, or $$ 2 U_4 = \frac{40}{56} U_1 = \frac{5}{7} U_1 $$
Subracting those two equations which express \$U_4\$ in \$U_1\$ from each other, gives \$ 0 = (3/x - 1 - \frac{5}{7}) U_1 \$, or
$$ x = 2 - \frac{1}{4} $$
Hence the equivalent resistance between terminals \$ 1 \$ and \$ 2 \$ equals \$ R_{12} = 17.50 \Omega \$. (How exciting...)
It's a two node circuit after suppressing the sources and connecting a test source between 1 and 2. Write the two node voltage equations and solve.
While the delta-star conversion will work I have a feeling most of us won't have it memorized.
Once you have shorted out the voltage source and open-circuited the current source, you apply a new source to terminals 1 and 2. For convenience, apply a 1 volt source and, through analysis, the current that flows is the inverse of the resistance between nodes 1 and 2: -
I can't see how I should redraw the circuit to see if it is a series or parallel circuit
That's because it doesn't consist of simple series or parallel combinations of resistors. You have to use circuit theories to determine the resistance.
EDIT to help OP: star delta transform calculator link: -
Or, just use microcap: -
Looks like 17.5 Ω resistance to me.
