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I am self-studying circuit analysis, and got the following question incorrect.

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I thought the answer was (a) since I know current is equal to voltage divided by resistance. I summed the resistors giving me $10$ ohms, and summed the voltages giving me $8$ volts. $8/10 = 0.8$ volts, and the current seems to be going from negative to positive, which is why I made it negative, hence $-0.8$ amps. The correct answer is $-0.2$ amps. Could someone explain why? I believe my summing of voltages was incorrect.

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You're right, your summing of voltages is incorrect. The voltage sources are in opposing polarities ('pushing' against each other), so the total voltage is actually $5-3 = 2$ V.

Using Kirchoff's laws, and starting from the position of the arrow, gives:

$-5V-6I+3V-4I=0$

$-2V-10I=0$

So $I = \frac{2V}{-10}$

The reasoning behind my choice of signs for the voltage drops is:

  • I follow the arrow around the circuit
  • When I come to a voltage source, the voltage is negative if I'm going towards the negative terminal and positive if I'm going towards the positive terminal
  • When I come to a voltage drop (the resistors), the voltage is negative if I'm going with the current, and positive if I'm going against the current.

So the current is 0.2A, but going in the direction opposite to the arrow, which makes sense because the voltage sources are in opposing polarities, and the 5V source will tend to push the current anti-clockwise.

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  • $\begingroup$ Thank you! I found this really helpful! $\endgroup$
    – Catdog
    Oct 17 '21 at 21:39
  • $\begingroup$ @Jim421616 Although you have taken time to answer the question you should not have done as this is a homework type question and you have provided the solution. $\endgroup$
    – Farcher
    Oct 17 '21 at 21:58
  • $\begingroup$ Good point. Should I edit? $\endgroup$
    – Jim421616
    Oct 17 '21 at 21:58

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