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

• subzero0137
In summary, the conversation discusses the application of relevant equations V=IR and KVL to determine the voltage and current across different resistors in a circuit with open and closed switches. The summary also mentions the use of R4/R1 = R3/R2 as a balancing condition for equal currents in R3 in both cases. The participants also speculate on the underlying principle or concept involved in the problem.
subzero0137

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.

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!

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?

Last edited:

## What is a resistor circuit?

A resistor circuit is a network of resistors connected in a specific arrangement. This arrangement determines the overall resistance of the circuit and how current flows through it.

## What is the purpose of a resistor in a circuit?

The purpose of a resistor in a circuit is to limit the flow of current and reduce the voltage in a circuit. This helps protect other components and ensures that the circuit operates within safe limits.

## How do you calculate the total resistance in a series circuit?

In a series circuit, the total resistance is equal to the sum of all individual resistances in the circuit. This can be calculated using the formula: Rtotal = R1 + R2 + R3 + ... + Rn

## How do you calculate the total resistance in a parallel circuit?

In a parallel circuit, the total resistance is the reciprocal of the sum of the reciprocals of each individual resistance. This can be calculated using the formula: Rtotal = 1/(1/R1 + 1/R2 + 1/R3 + ... + 1/Rn)

## What is the relationship between voltage, current, and resistance in a circuit?

According to Ohm's Law, the voltage in a circuit is equal to the current multiplied by the resistance (V=IR). This means that as resistance increases, the voltage will also increase, and as resistance decreases, the voltage will decrease. The current in a circuit is directly proportional to the voltage and inversely proportional to the resistance.

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