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I came across this problem and I'm trying to really understand it.

Problem setup

  1. VB is the voltage drop across a particular element, it is the change in voltage from node a to node b. Find the value of VB in volts and enter it in the box below.

  2. The expression Vab indicates the voltage referenced from node b to node a. In other words, imagine Vab being identified in the diagram as having a "-" at node b and a "+" at node a. For the circuit above, find Vab and enter it into the space below without units.

  3. For the same circuit above, find VC and enter it into the space below without units. (VC is the voltage across the top element, not the voltage at the point c in the circuit.)

The answers are:

  • 1) 3
  • 2) -3
  • 3) -2

Now, I know that since VA is 3V so do VB because they are in parallel. but why is not VC + VD= 3V since they are also in parallel to VA

what is throwing me off is that if VA=3V that means the voltage at node b is 3V now the voltage at node a is 1V, that would mean that the voltage at node b minus the voltage at node a is equal to 2, (i.e. Vb-Va=3V-1V=2V=VB) but it is 3 instead.

The way I solved for VC is just intuitive, subtracted VA-VD and then multiplied by a negative sign because the negative sign is first (assuming that my current flows clockwise through the whole circuit), why can I do this fro VB?

Also why is Vab negative if voltage is dropping and not raising?

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Voltage is the difference in electric potencial between two points.

The electric potencial of point a is 1V. Your mistake was to assume that the electric potencial in point b must be 3V because of the voltage drop of 3V. Actually the electric potencial in point b is 4V.

VA = Vb - Va
3V = Vb - 1V
Vb = 3V + 1V
Vb = 4V

In words: The electric potencial in point b is 3V higher than the electric potencial in point a. Resulting in a voltage drop over the Resistor VA of 3V.

In the same way you can calculate the electric potencial in point c.

VD = Vc - Va
1V = Vc - 1V
Vc = 1V + 1V
Vc = 2V

And last the voltage drop VC:

VC = Vc - Vb
VC = 2V - 4V
VC = -2V

Quick Scetch:

enter image description here

Hope that helps.

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  • \$\begingroup\$ Makes a lot of since there is only 1 thing still confusing: If voltage rose from 1V at a to 4V at b, why is not rising from b to c? and instead is dropping from 4V to 2V? \$\endgroup\$ – Marco Castro Aug 21 '18 at 9:05
  • \$\begingroup\$ You can calculate the electric potencial in point c. It must be 1V higher than the electric potencial in point a, because of the voltage drop VD (1V). So the electric potencial in point b = 4V and the electric potencial in point c = 2V. So over the resistor VC must be a voltage drop of 2V from point b to point c. It is -2 because of the given voltage drop orientation. \$\endgroup\$ – user196456 Aug 21 '18 at 10:10
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You can treat the voltage as a height. https://www.youtube.com/watch?v=ZSrzvoJcaJ8&g

To measure the voltage we need two points in the space. One of this point is treated as a reference point. We have a very similar situation when we try to measure a height of an object. We need a reference point. The most common reference point is "above mean sea level". But when you measure the height of the table in your house the floor now becomes your reference point.

enter image description here

enter image description here

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  1. \$V_{C} + V_{D}\$ is indeed equal to 3V, if you take into account the reverse polarity of \$V_{C}\$. Change the polarity across \$V_{C}\$ component. And then sum up the values of \$V_{C}\$and \$V_{D}\$, i.e, 2 + 1 = 3V.
  2. \$V_{A}\$ = 3V refers not to the potential at node b, but rather the potential difference across the component. Thus, if potential at node a equals 1V, then potential at node b equals \$V_{a} + V_{A} = 1 + 3 = 4V\$. Please get this concept properly.
  3. Solving for any of the variables asked above is best done by writing Kirchoff's Voltage Law equations for the respective loop. You can find the value of \$V_{B}\$ by considering the loop a->b->c->a and you will get the following equation:

$$V_{B} + V_{C} - V_{D} = 0$$ $$ V_{B}-2-1 = 0 \implies V_{B} = 3V$$

  1. According to what is given in the question, \$V_{ab} = V_{a} - V_{b}\$. Since node a is at a lower potential compared to node b, you end up with a negative value.
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