2
\$\begingroup\$

I am a student in 10th grade and our teacher gave us this schematic to work with (Remade with circuit lab):

schematic

simulate this circuit – Schematic created using CircuitLab

He asked us to find the equivalent resistance of the circuit. Based on my calculations, 24 Ohms should be the right answer. Apparently my teacher disagreed and told us it was around 2.5 Ohms or so.

I don't know which answer is the correct one, but I am fairly convinced that mine is the most correct one. Since even if my calculations of the series-parallel resistors are wrong, the minimum resistance is going to be 20 Ohms (R1 + R7).

\$\endgroup\$
  • 7
    \$\begingroup\$ When asking for eq. resistance one should tell between which points. Assuming as seen by power supply, it is definitely can't be less than 20 Ohm as there are two 10 Ohm resistors in series. I would say you might have misunderstood the question. \$\endgroup\$ – Eugene Sh. Apr 24 '18 at 20:50
  • 2
    \$\begingroup\$ Ask your teacher, "Pretend that R6 is zero ohms. What is the equivalent resistance?" If he answers anything other than 20 ohms, you know you've got a problem. \$\endgroup\$ – WhatRoughBeast Apr 24 '18 at 20:52
  • 3
    \$\begingroup\$ I get about 2.6 Ohms looking from the right into the cirtuit, shorting the supply. \$\endgroup\$ – Oldfart Apr 24 '18 at 20:56
  • \$\begingroup\$ @Samuel, no, if ideal VOLTAGE source, impedance is usually 0. \$\endgroup\$ – Andrés Apr 24 '18 at 21:05
  • 1
    \$\begingroup\$ Equivalent resistance seen by the source? Or between some other two nodes? If it's what's seen by the source, you know it can't be less than 20, because you have R1, R7, and "the rest of the network" in series. \$\endgroup\$ – The Photon Apr 24 '18 at 21:22
0
\$\begingroup\$

The answer is 26.25 ohms. Verified with a spice modelling program.

\$\endgroup\$
  • 3
    \$\begingroup\$ We don't give the answer homework questions on EE.SE. Instead give pointers to allow the question asker to figure it out for themselves. \$\endgroup\$ – Tom Carpenter Apr 24 '18 at 23:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.