Question about voltage drops in closed circuit

[virtuoso735]

Can someone help me with this problem? I'm struggling to understand voltage drops in a closed circuit. It seems to me that the sum of the voltage drops in a circuit should equal the total voltage that you start with. However, the problems that I'm doing don't seem that way.

For example, in BR Physics Book II Section IX on page 196, there is a diagram on the bottom that I don't understand. I spent half an hour trying to draw a picture and uploading it, but it comes out very unclear since I used paint and don't know how to make it clearer, so I hope someone who has the book can look it up and explain it to me.

I don't understand why the voltage drops for the two resistors in parallel in the loop add up only to -3. I understand that they should add up to -3, since the sum for all the voltage drops should equal -6, which makes sense, but the voltage drop V2 = -I2R2 = -(1 A)(3 ohms) = -3 V, right? And then the bottom resistor has voltage drop V3 = -I3R3 = -(0.5 A)(6 ohms) = -3 V. But the book shows that the total voltage drop for the two only equals -3 V. Why is this?

---

Members don't see this ad.

 

[SN2ed]

Thread moved. These types of questions belong in the Study Q&A forum.

 

[MT Headed]

I'm guessing that you are going through one resistor in the forward direction (I.e. with the current) and the other resistor in the backwards direction. so the voltage would drop through the first resistor, and then be regained as you go backwards through the second, thereby proving that the voltage drop in (any) closed loop is zero..

And really that's a better way to think about voltage gains and drops. Any path that gets you back to where you started must have a total drop of zero.

 

[Ssina]

from the kirchhoff's rules (on pg. 188):
Parallel: Voltage the same, and current adds up
Series: Current the same, and voltage drops add

so since R2 and R3 are parallel the voltage drop will be -3 not only for each, but for their combination (you can see it this way also: Req is 2 and the current across both is the sum 1+0.5= 1.5.... so the Voltage drop across the equivalent resistor is V=IR=1.5 x 2= 3 volts again)

 

[Ssina]

I'm guessing that you are going through one resistor in the forward direction (I.e. with the current) and the other resistor in the backwards direction. so the voltage would drop through the first resistor, and then be regained as you go backwards through the second, thereby proving that the voltage drop in (any) closed loop is zero..

And really that's a better way to think about voltage gains and drops. Any path that gets you back to where you started must have a total drop of zero.

if you follow the current from around the circuite, it goes from the higher potential side of the batter (longer line) toward the lower potential (shorter line).... so it'll pass through R1 (lose 3 volts) and then you have the option of going through R2 or R3 and we have only 3 volts to lose (6-3=3volts)... so either way you go, the voltage drop is 3V