-3
\$\begingroup\$

A battery of EMF 8.0 V and internal resistance r = 1.0 – is connected to an external circuit as shown:

I need to find out the total resistence and my problem is that I dont know which resistors consider parallel and which are in series as there is connection for 3ohm resistor after 1st resistor which confuses me. Answer I got is that the total resistance is 15ohms could anyone confirm it? Or point out where I went wrong?

\$\endgroup\$
  • \$\begingroup\$ I'm curious how you found 15\$\Omega\$ \$\endgroup\$ – jippie Mar 3 '13 at 22:32
  • 2
    \$\begingroup\$ Homework, per chance? \$\endgroup\$ – Keelan Mar 4 '13 at 6:44
  • 2
    \$\begingroup\$ Even without doing the complete calculation you should know 15 ohm can't be right: the left loop totals to only 9 ohm, and the right part can only decrease that, never increase. \$\endgroup\$ – amadeus Dec 18 '13 at 11:04
5
\$\begingroup\$

Start with the part that's easy, and work your way out. 6 ohms in parallel with 12 ohms is 4 ohms. 4 ohms in series with 2 ohms is 6 ohms. 6 ohms in parallel with 3 ohms is 2 ohms. 1 ohm in series with 2 ohms in series with 4 ohms in series with 1 ohm is 8 ohms.

\$\endgroup\$
  • \$\begingroup\$ so does last resistor "r" in this case have anything to do with total resistance of rest of resistors? \$\endgroup\$ – corkalom Mar 3 '13 at 19:25
  • 2
    \$\begingroup\$ "r" is the final 1 ohm. Yes, it affects the total resistance of the circuit. \$\endgroup\$ – Pete Becker Mar 3 '13 at 22:19
  • 2
    \$\begingroup\$ If r is the internal resistance of the voltage source, then it does not count to the circuit's total resistance, it is part of the voltage source. It does however have to be taken into account when trying to calculate the current in the circuit. This can be a discussion point, if you are in doubt during an exam then the safe choice is to explain and calculate both scenarios for the circuit. My bet is that most EE's will argue that r is not part of the circuit, but is part of the source. \$\endgroup\$ – jippie Mar 3 '13 at 22:24
2
\$\begingroup\$

Assuming that the small r is to be neglected 15 Ohms is NOT correct. To point out where you went wrong you must first explain how you got to your answer!

The easiest way to solve this is to cut the problem to pieces, and solve them step by step.

First (in your mind) replace the parallel 6 and 12 Ohm resistors with their combined equivalent, then combine that with the 2 ohm resistor that is in series.

See if you can find the next step for yourself.

\$\endgroup\$
  • \$\begingroup\$ corkalom specified the small r has a resistance of 1.0Ω \$\endgroup\$ – Huey Jan 7 '15 at 13:38
1
\$\begingroup\$

Always start from Opposite side of EMF Source. And also, resolve parallel connections first!

So, 6 ohm and 12 ohm in Parallel.

Resultant is in series with 2 ohm

Resultant is in parallel with 3 ohm

And then All are in Series!

\$\endgroup\$
0
\$\begingroup\$

Other posters have covered the general procedure: Look for any cases where you have two or more resistors directly in parallel with each other, or directly in series, and replace each of those with the calculated equivalent single resistor. Those new single resistors will make simple parallel or series combos, so apply procedure again until done.

But here's the trick to doing it in your head. Of course the series combos are easy -- just addition. At first glance the parallel combos seem difficult since the calculation involves reciprocals or equivalent mental gymnastics. However, you can often apply this trick:

a) N parallel resistors each of value R have equivalent resistance R/N. So 3 resistors of 100 ohms in parallel equal 100/3 = 33.3 ohms.

b) For parallel resistors that don't happen to be of the same value, visualize each of the resistors as parallel combos of resistors of some common multiple!

In this example, the first parallel pair to attack is the 6 || 12 combo in the right most branch. OK, the 6 ohm resistor is equivalent to 12 || 12, so overall what we have is like 3 resistors of 12 ohms in parallel = 12/3 = 4.

So now we have the right branch as 4+2 = 6 ohms, parallel to the middle branch of 3 ohms. 3 is like 6 || 6, so the effect of the middle and right branches is 3 resistors of 6 ohms in parallel, = 6/3 = 2.

And so on.

\$\endgroup\$
0
\$\begingroup\$

Start at the rightmost part of the circuit. Observe that the resistors 6 ohms and 12 ohms are in parallel, so you need to combine them. once your are done, 6//12 (// means parallel) is in series with 2 ohms resistor, so you can just add them. Then again, 6//12+2 (6//12 means 6 ohms is parallel with 12 ohms and '+' means both are in series with 2 ohms resistor) is in parallel with 3 ohms resistor; just repeat what you did in the 6//12 ohms resistors. After that, the three resistors 4 ohms, the combined resistors [6//12) + 2 ]//3 ohms, and the internal resistance 1 ohm is now in series with each other. That means you can just add them to get the total resistance of the circuit (your answer should be 8 ohms).

\$\endgroup\$
0
\$\begingroup\$
  1. (6 // 12) + 2 = 6

  2. 6 // 3 = 2

  3. 4 + 2 + 1 = 7

[POWER = IE = 8V/7 OHMS = I =1.14429 X E 8V = 9.15 WATTS

*PARALLEL(//)

\$\endgroup\$

protected by Dave Tweed May 20 '16 at 13:09

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

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