A linear circuit is shown below in Figure 1.

Figure 1
The elements in this circuit have the following values: R1=100Ω, R2=200Ω, R3=350Ω, V=5V, and I=0.004A.

Determine the potentials drops vA and vB across resistors R1 and R2 respectively.

What is the value of voltage vA in Volts?

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What is the value of voltage vB in Volts?

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Now let us determine if the answers you came up with satisfy the laws of physics.

What is the power (in Watts) dissipated in resistor R1?

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What is the power (in Watts) dissipated in resistor R2?

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What is the power (in Watts) dissipated in resistor R3?

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What is the power (in Watts) coming out of the voltage source V?

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What is the power (in Watts) coming out of the current source I?

How are the resistors connected?

To solve this problem, we need to analyze the given linear circuit and use Ohm's law and power formulas.

1. First, let's determine the potential drop vA across resistor R1.
The potential drop across a resistor is given by Ohm's law: v = i * R, where v is the potential drop, i is the current passing through the resistor, and R is the resistance.
In this case, the current passing through R1 is I = 0.004A, and the resistance is R1 = 100Ω. So, vA = I * R1 = 0.004A * 100Ω = 0.4V.

2. Next, let's find the potential drop vB across resistor R2.
Again, applying Ohm's law, vB = I * R2 = 0.004A * 200Ω = 0.8V.

3. Now, let's check if our answers satisfy the laws of physics.
- To determine the power dissipated in a resistor, we can use the formula: P = v^2 / R, where P is the power, v is the potential drop, and R is the resistance.
The power dissipated in R1 is P1 = vA^2 / R1 = (0.4V)^2 / 100Ω = 0.0016W (or 1.6mW).
The power dissipated in R2 is P2 = vB^2 / R2 = (0.8V)^2 / 200Ω = 0.0032W (or 3.2mW).
The power dissipated in R3 cannot be determined without additional information, such as the current passing through it.
- To find the power coming out of a voltage source, we can use the formula: P = v * i, where P is the power, v is the voltage, and i is the current.
The power coming out of the voltage source V is PV = V * I = 5V * 0.004A = 0.02W (or 20mW).
- To find the power coming out of a current source, we need to know the voltage across it, which is not provided in the problem statement. Therefore, we cannot determine the power coming out of the current source I.

So, to summarize:
- The potential drop vA across resistor R1 is 0.4V.
- The potential drop vB across resistor R2 is 0.8V.
- The power dissipated in resistor R1 is 0.0016W (or 1.6mW).
- The power dissipated in resistor R2 is 0.0032W (or 3.2mW).
- The power dissipated in resistor R3 is unknown.
- The power coming out of the voltage source V is 0.02W (or 20mW).
- The power coming out of the current source I cannot be determined without additional information.