Circuit concept confusion

[AA|FCB|DOC]

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I just had a question about the attached problem. To be exact, I actually understand the problem and the answer choices just fine. I have an issue with something else I was thinking about while looking at the problem. Since we are adding another resistor parallel when the switch is closed, then we are decreasing the overall resistance of the circuit and therefore increasing the overall current that goes through the circuit. In addition, since the resistor is added in parallel, the voltage drop of R1 will stay the same as it was before the switch was closed. I get this too. But here is where I am confused. If the overall current of the circuit increases then the current through the unlabeled resistor (the 0.1 ohm one in series with the two parallel ones) should increase in voltage drop shouldn't it? And if it does then how could the voltage of the parallel resistors not change since the overall voltage will be (V1=V2) + V unlabeled resistor = V battery. Thanks in advance for any help.

 

[milski]

The voltage drop across R1 will not be the same. You are correct that the current through the unlabeled resistor will increase. As a result, the voltage drop across the unlabeled resistor will also increase - there is no way around Ohm's law, it is always true. Since (V1=V2)+Vunl=Vbat, V1 and V2 have to decrease.

 

[AA|FCB|DOC]

yea I was thinking the exact same way you were, but if we are correct then would it not make this problem incorrect? I know choice C says the voltage drop gets larger and therefore that is still an incorrect answer, but when I looked at the explanation to see why it is incorrect it says that since the other resistor is getting added parallel when the switch is closed, the voltage drop stays the same. You and I both seem to think that is gets smaller.

 

[milski]

We are right, the explanation is incorrect.

When both switches are closed, the equivalent resistance of R1 and R2 is 1/(1/0.1+1/10)=.099 &#937;, the resistance of the circuit is 0.199 &#937; and the current is V/0.199. Then the voltage across the unnamed resistor is V/0.199*0.1=1/1.99 V. The voltage across each R1 and R2 is V-1/1.99V=0.99/1.99V < .5V

When only switch one is closed, the total resistance is 0.2 &#937;, the current is V/0.2 and the voltage drop across R1 is 0.1/0.2V=0.5V

 

[AA|FCB|DOC]

haha ok thanks a lot milski. Just trying to make sure we are still sane 🙂