Solving Unknown Resistance in Series Circuit

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The discussion revolves around solving for an unknown resistance R in a circuit where a 7-ohm resistor and R are in parallel, connected in series with a 12-ohm resistor and a battery. The key point is that the current through the 7-ohm resistor remains constant in both configurations, leading to the conclusion that the voltage drop across the 7-ohm resistor must be the same in both cases. Participants suggest setting up equations for the voltage in each configuration and equating them to find R. Despite attempts to simplify the equations, the user encounters an identity (1=1), indicating a need for a different approach. The solution lies in recognizing that the voltage drops across the known resistors must equal the voltage drop across the 12-ohm resistor in the first circuit.
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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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