How Does Internal Resistance Affect Terminal Voltage in a Circuit?

[mer584]

Homework Statement

Three equal resisters are connected to a battery. If the emf of the battery is 15V, what is its terminal voltage when the switch is closed if the internal resistance is 0.5 ohms and R= 5.5ohms? What is the terminal voltage when the switch is open.

Homework Equations

V=IR; V terminal = E- Ir

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The Attempt at a Solution

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So I'm not sure if you need to apply Kirchhoffs rules here or not. It seems simple enough that you might not have to?
I first tried treating everything vertically as parallel and then adding it to the horizontal R as though it were in series

Then I tried adding everything in the first box as though it were in series and adding that to the outside R as though they were in parallel with each other.

In both cases I then used the equivilant value for R and used it with the emf to find the terminal voltage. Both cases gave me voltages in the high 13's, but the answer is 14.1V can anyone start me off the right way?

 

[mer584]

Nevermind, I eventually got it to come out to the right answer

 

[Moetasim]

I can provide a response to this content by explaining the concept of terminal voltage and how it relates to circuits.

Terminal voltage is the potential difference between the terminals of a battery or any other source of electrical energy. In this case, we have a battery with an emf (electromotive force) of 15V. This means that the battery can provide a potential difference of 15V to the circuit. However, when the switch is closed, the internal resistance of the battery (0.5 ohms) will cause a drop in voltage, resulting in a lower terminal voltage.

To calculate the terminal voltage when the switch is closed, we can use the equation V terminal = E - Ir, where V terminal is the terminal voltage, E is the emf, I is the current flowing through the circuit, and r is the internal resistance. In this case, we have three resistors of equal value (5.5 ohms each) connected in parallel to the battery. This means that the total resistance of the circuit is 5.5/3 = 1.83 ohms.

To find the current flowing through the circuit, we can use Ohm's law, V=IR, where V is the voltage drop across the resistor (which is equal to the emf of the battery), I is the current, and R is the resistance. So, I = V/R = 15/1.83 = 8.2A.

Now, we can plug in these values in the equation for terminal voltage: V terminal = 15V - (8.2A * 0.5 ohms) = 14.1V. This is the correct answer.

When the switch is open, there is no current flowing through the circuit, so the voltage drop across the internal resistance is zero. Therefore, the terminal voltage will be equal to the emf of the battery, which is 15V.

In summary, the terminal voltage of a battery depends on the emf of the battery and the current flowing through the circuit, taking into account the internal resistance of the battery. When the switch is closed, the terminal voltage will be lower due to the voltage drop across the internal resistance. When the switch is open, the terminal voltage will be equal to the emf of the battery.