Switch in parallel resistor circuit

[Hemorrage]

Can someone explain TBR physics p220 #46? I'm confused about the concept of a switch in a parallel circuit. The answer says that the voltage across the parallel circuit is the same regardless of whether its connected or not but that doesn't make sense to me. I thought that voltage only arises when theres flow of charges from higher PE to lower PE (hence the definition of Voltage).

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[milski]

The potential is a difference in the strength of the electrical field. It tells you how hard it will be for a charge to move between the two points. It does not tell you that there will be a charge to be moved or that there is a way for the charge to get from point A to point B. In a way, it's like the distance between two cities - you get an idea how far they are from each other but you don't know if there are people willing to go from city A to city or if there's a road between the two cities at all.

In a related note, keep in mind that while potential is somewhat similar in its behavior to potential energy, it is not the same thing. In a mechanical analogy, it will be the equivalent of the height of an object, not to the PE of that object.

 

[dudewheresmymd]

The concept behind this is that the voltage drop across each branch in a parallel circuit is the same. But current is different among the branches. Take the two situations below. Just by closing the switch, the voltage of the battery does not change 100volts must be supplied across each branch initially. If this is true the current through R1 both before and after, must stay the same because the resistance of R1 does not change. V=IR, V(same before and after)/R(same before and after) = I(same before and after).

What does change, however, is the current through the middle branch as can be seen below: For clarity sake, you can try this math approach below just so you understand it, but definitely will waste time on test day....try to understand the above concept through the math. Hope this helps.

Voltage of battery=100v, each resistor = 5 ohms

Before:
Top branch: Voltage across first resistor R1 = 100v, R1=5 ohms, R2, 5ohms, Itop=100Vtotal/10ohms=10A
Bottom branch: Voltage across first resistor R3=100v, R3=5ohms R4=5ohms, Ibottom=100Vtotal/10ohms=10A
Req= (10*10)/(10+10)=100/20=5ohms
I total= 100V/5= 20A=Itop+Ibottom


After:
Top branch: Voltage across first resistor R1= 100v, R1=5 ohms, R2=5ohms, 100Vtotal/10ohms=10A, Itop=100Vtotal/10ohms=10A
Middle branch: Voltage across resistor R5= 100V, R5=5ohms, 100V/5ohms=20A, Imiddle=100Vtotal/5ohms=20A
Bottom branch: Voltage across branch R3= 100v, R3=5 ohms, R4, 5ohms, 100Vtotal/10ohms=10A, Ibottom=100Vtotal/10ohms=10A
Req=(5*5)/(5+5)=2.5ohms=Itop+Imiddle+Ibottom
Votal/Req= 100V/2.5= 40A

Thus, the current across branches 1 and 3 do not change, but when you add the middle branch in, the total equivalent resistance decreases, which leads to an increase in the total current, manifested through an increase in 20A being supplied only to branch 2, I1 and I3 stay constant.

Cheers.

 

[Hemorrage]

The concept behind this is that the voltage drop across each branch in a parallel circuit is the same. But current is different among the branches. Take the two situations below. Just by closing the switch, the voltage of the battery does not change 100volts must be supplied across each branch initially. If this is true the current through R1 both before and after, must stay the same because the resistance of R1 does not change. V=IR, V(same before and after)/R(same before and after) = I(same before and after).

What does change, however, is the current through the middle branch as can be seen below: For clarity sake, you can try this math approach below just so you understand it, but definitely will waste time on test day....try to understand the above concept through the math. Hope this helps.

Voltage of battery=100v, each resistor = 5 ohms

Before:
Top branch: Voltage across first resistor R1 = 100v, R1=5 ohms, R2, 5ohms, Itop=100Vtotal/10ohms=10A
Bottom branch: Voltage across first resistor R3=100v, R3=5ohms R4=5ohms, Ibottom=100Vtotal/10ohms=10A
Req= (10*10)/(10+10)=100/20=5ohms
I total= 100V/5= 20A=Itop+Ibottom


After:
Top branch: Voltage across first resistor R1= 100v, R1=5 ohms, R2=5ohms, 100Vtotal/10ohms=10A, Itop=100Vtotal/10ohms=10A
Middle branch: Voltage across resistor R5= 100V, R5=5ohms, 100V/5ohms=20A, Imiddle=100Vtotal/5ohms=20A
Bottom branch: Voltage across branch R3= 100v, R3=5 ohms, R4, 5ohms, 100Vtotal/10ohms=10A, Ibottom=100Vtotal/10ohms=10A
Req=(5*5)/(5+5)=2.5ohms=Itop+Imiddle+Ibottom
Votal/Req= 100V/2.5= 40A

Thus, the current across branches 1 and 3 do not change, but when you add the middle branch in, the total equivalent resistance decreases, which leads to an increase in the total current, manifested through an increase in 20A being supplied only to branch 2, I1 and I3 stay constant.

Cheers.

I see! So basically the equivalent resistance will increase, however, the resistance of the permanently connected resistors will not change?

Also, what if it was a series circuit with 2 resistors. I know each resistor "subtracts" from the potential (20V for example), the voltage will be 0 when the switch isn't connected right? Since there isn't any current?

 

[dudewheresmymd]

If only a series circuit with 2 resistors, the current is the same through both resistors, (2 5 ohm reistors) 20V battery, Itotal=Vtotal/Rtotal= 20/10=2A through each. Because it's series, the Itotal=the current for both resistors (by definition). However, the voltage is not the same in series across each resistor, there is a voltage drop as you alluded to each time you go through a resistor in series, in this example, V=IR, 2*5=10volts. After current passes through first resistor, voltage drops to (20-10)=10V, after going through 2nd resistor (V=IR) 2A*5ohms=10V, (10V-10V=0), voltage drops down to 0, or as TBR would show (20V+-10V+-10V=0) (Kirchhoff's 2nd law). However, if you now look at the 2nd resistor now, and do V=IR, (the voltage entering the 2nd resistor is 10V (NOT THE ORIGINAL 20 V)/5ohms= 2A=Itotal from above)...everything matches! As long as your calculations match, you know you are doing the calculations right, just as a tip.

Not sure what you mean "the voltage will be 0 when the switch isn't connected)... if the switch isn't connected there is still a voltage in the battery, but that potential difference CANNOT drive the flow of electrons b/c the circuit is incomplete. you need "some sort of resistance or load" be it in a closed loop or a resistor, to complete the circuit, and thus for the potential difference of the battery to pump electrons through the circuit to generate a current! V/R= I, I only exists if there is resistance, if the circuit is incomplete (ie there is an open switch) or if there is no circuit, just a battery, there is still the voltage between the terminals of the battery (referred to as open circuit voltage), but b/c there is no complete circuit (resistance), there is no current. Think about a 120V A/C plug, it can still shock you if nothing is plugged inside of it (therefore it has potential (voltage) even if there is no circuit (something plugged into it).

 

[getsome111]

in the initial question would this relationship still hold if r5 was of extremely low resistance? Im having problems understanding how there isnt a redistribution of current in all cases (including short circuits)