How do I calculate voltage drops and currents in a resistor circuit problem?

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To calculate voltage drops and currents in a resistor circuit, use the formula V=IR. When switches are open, the total current is 1.5 A, calculated from the series resistance. With the switches closed, R2 is shorted, and R3 maintains the same current, leading to a voltage drop of 4.5 V across it. Applying Kirchhoff's Voltage Law (KVL) helps determine the voltage across R1 and the current through it, confirming the calculations. The discussion also touches on a potential relationship between resistances in the circuit, suggesting a balancing condition for currents.
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Relevant equations:
V=IR

Attempt at solution:

When the switches are open, R1, R2 and R3 are in series so the current in that case would be 12 V/ (2 ohms + 3 ohms + 3 ohms) = 1.5 A. When the switches are closed, the current across R3 is still 1.5 A, so the voltage drop across R3 is 1.5 A * 3 ohms = 4.5 V. I don't know how to calculate the voltage drops across other resistors and the currents through them. How do I proceed? I know we can ignore R2 when the switches are closed since it gets shorted, and I know R and R3 are in parallel when the switches are closed.
 
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Your reasoning so far is correct.
subzero0137 said:
so the voltage drop across R3 is 1.5 A * 3 ohms = 4.5 V.
Yes.
So, what is the voltage across R1? Apply KVL to the loop containing ε, R3 and R1.
 
cnh1995 said:
Your reasoning so far is correct.

Yes.
So, what is the voltage across R1? Apply KVL to the loop containing ε, R3 and R1.

ε - (I*R1) - (1.5A *R3) = 12 V - (I*R1) - (1.5*3) = 0 therefore I*R1 = 7.5 V. The current across R1 is 7.5/3 = 2.5 A, therefore the current across R = 1 A. Since the voltage drop across R and R3 is the same (4.5 V), the resistance of R is 4.5/1 = 4.5 ohms. Is that correct?
 
subzero0137 said:
ε - (I*R1) - (1.5A *R3) = 12 V - (I*R1) - (1.5*3) = 0 therefore I*R1 = 7.5 V. The current across R1 is 7.5/3 = 2.5 A, therefore the current across R = 1 A. Since the voltage drop across R and R3 is the same (4.5 V), the resistance of R is 4.5/1 = 4.5 ohms. Is that correct?
It is perfect! :smile:
 
Complicated simplification?
If we call the unknown resistance R, R4, then does everybody agree that this problem gives in general

R4/R1 = R3/R2
as the condition for currents in R3 equal for both cases?
A formula (recalling a bridge balancing condition) that seems to be 'suggestive' - but I don't know what it suggests. Can the circuits be switched around somehow to make it obvious what the answer should be? Is there some known principle involved? Duality or something?
 
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