1
$\begingroup$

Problem:

Calculate the equivalent resistance, $R_\text{eq} ,$ of this circuit:

$\hspace{50px}$.


My solution attempt

  1. Тhe $12 \, \Omega$ resistor and the $6\, \Omega$ resistor are in parallel, so$$ R = {\left(\frac{1}{12\,\Omega} + \frac{1}{6\,\Omega} \right)}^{-1} = 4\,\Omega \,, \tag{1} $$

    reducing the circuit diagram to:
    $\hspace{50px}$.

  2. The $4 \,\Omega$ resistor and the $12 \,\Omega$ resistor are in series, so$$ R = 12\,\Omega + 4\,\Omega = 16\,\Omega \,, \tag{2} $$

    reducing the circuit diagram to:
    $\hspace{50px}$.

I am not sure of the answer, and I am so confused right now.

I am totally new to circuits and any help will be much appreciated.

| cite | improve this question | | | | |
$\endgroup$
  • $\begingroup$ Is the top circuit diagram exactly as it appears in the book? $\endgroup$ – Bob D Aug 13 '19 at 20:42
  • $\begingroup$ @BobD Yep , it's exactly the same $\endgroup$ – AmirWG Aug 13 '19 at 20:45
  • 2
    $\begingroup$ The reason I ask is the 12 ohm resistor on the left is short-circuited by the wire and can be replaced by a wire. Then you have a parallel combination of 6 ohm and 12 ohms and that would give you an answer different than the book. $\endgroup$ – Bob D Aug 13 '19 at 20:49
  • 1
    $\begingroup$ I see 4 Ohms also $\endgroup$ – Dale Aug 13 '19 at 20:50
  • $\begingroup$ how do you tell if a resistor is short-circuited or not ? $\endgroup$ – AmirWG Aug 13 '19 at 20:51
2
$\begingroup$

We are not supposed to provide solutions to homework and exercise questions, only guidance.

So here is some guidance;

The top circuit diagram below is your top diagram. The bottom diagram is equivalent to the top diagram, as long as all the "wires" shown are considered ideal (that is, zero resistance).

So given the bottom diagram, what should the answer be, regardless of what the "book" says.

Hope this helps

enter image description here

| cite | improve this answer | | | | |
$\endgroup$

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