Solving Unknown Resistance in Series Circuit

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Homework Help Overview

The problem involves analyzing a circuit with a known resistor and an unknown resistance in two different configurations: a parallel connection with a series resistor and a series-only connection. The goal is to find the unknown resistance while ensuring that the current through a specific resistor remains constant in both setups.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the need to equate the potential difference across the known resistor in both circuit configurations. Some express confusion about the conditions under which the voltage drop remains the same, while others attempt to derive equations based on current and voltage relationships.

Discussion Status

The discussion is ongoing, with participants exploring different equations and relationships between current and voltage in the circuits. Some have provided guidance on how to set up the equations, while others are questioning the assumptions made regarding voltage drops and current flow.

Contextual Notes

Participants note that the problem may lead to identities, suggesting that there may be missing information or assumptions that need to be clarified to progress further.

syang9
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This problem seems pretty straightforward.. and yet I haven't made any significant progress at all.

"A parallel combination of an 7 resistor and an unknown resistor R is connected in series with a 12 resistor and a battery. This circuit is then disassembled and the three resistors are then connected in series with each other and the same battery. In both arrangements, the current through the 7 resistor is the same. What is the unknown resistance R?"

So I drew the circuit out, and I concluded that for the last statement to be true, the equivalent resistances must be the same (since every resistor in a series circuit feels the same current, by virtue of being in series.) So I wrote

[(1/7) + (1/R)]^-1 + 12 = 7 + 12 + R

Unfortunately this equation is woefully uninformative, as when it is simplified, it yields 1 = 1. Could I have a hint or two from the masters?

Thanks,

Stephen
 
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You have to achieve the same potential difference across the 7 ohm resistor in both cases. Get an equation that expresses this voltage in terms of the unknown R for each circuit, and set them equal.
 
Why would the potential drop be the same for the 7 ohm resistor in both resistors? In series, wouldn't the voltage drop depend on where the 7 ohm resistor is connected?
 
syang9 said:
Why would the potential drop be the same for the 7 ohm resistor in both resistors? In series, wouldn't the voltage drop depend on where the 7 ohm resistor is connected?
The voltage drop on a resistor is RI. Since the problem states that the current is the same in both arrangements, the voltage drps must be equal, as OlderDan said.
 
So, here's what I have..

for the first circuit,

V = I1(7Ω) = I2(R); I1 + I2 = Itot

for the second circuit,

V = (7Ω + R)*Itot

so we have I2(R) = (7Ω + R)*Itot; R = Itot (7Ω) / I2. The thing is, Itot is defined in terms of the resistance and the voltage drop.. so I keep getting an identity (i.e., 1=1). So I really don't know what to do.. could you guys help me out?
 
syang9 said:
So, here's what I have..

for the first circuit,

V = I1(7Ω) = I2(R); I1 + I2 = Itot

for the second circuit,

V = (7Ω + R)*Itot

so we have I2(R) = (7Ω + R)*Itot; R = Itot (7Ω) / I2. The thing is, Itot is defined in terms of the resistance and the voltage drop.. so I keep getting an identity (i.e., 1=1). So I really don't know what to do.. could you guys help me out?
No, for the second circuit V = 7I1, just as for the first. The difference is that or the second circuit the current I1 passes also through the series connection of the two other resistors.
Since the battery is the same, the voltage drop on the two known resistors must be equal to the voltage drop in the 12 ohm resistor for the first circuit.
 

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