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- Thread starter apchemstudent
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In summary, the maximum value of R for a circuit to function is when it is equal to the internal resistance. This is because the potential drop across the resistor must equal the potential provided by the battery. Calculus is used to determine this and it is shown that at maximum power, the load resistance equals the internal resistance.

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apchemstudent

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anyone have any idea as to how the answer was reached?

- #3

asrodan

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The potential of the battery and the current produced are both constant.

Since the only components are the battery and the resistor, the potential drop across the resistor must equal the potential provided by the battery. So the maximum value of R for which the circuit would "function" is when it is equal to the internal resistance.

It = IR where t = internal resistance.

- #4

GCT

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From what I recall recently

[tex]V_{battery} = emf-IR_{internal}[/tex]

assuming that the "load" resistor is [tex]R_{eq}[/tex], (I forgot some of the terminologies), than

[tex]P=I^{2}R[/tex], where the [itex]I=EMF/(R+R_{internal})[/itex]

thus

[tex]P= \frac {(EMF^{2}(R))}{(R+R_{internal})^{2}} [/tex]

Try observing the trend for an infinitely large value of R and infinitely small value.

I'm pretty new to this also, so I'm not absolutely certain

- #5

Delta

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[tex]P = \frac{E^2R}{(R+r)^2}[/tex] as shown above.

where P is the power, R is the load, r is the internal resistance and E is the battery EMF.

We are trying to find the point at which P is a maximum by varying the load R, therefore we involve a bit of calculus with respect to R.

We need the quotient rule.

[tex]\frac{d}{dx}(\frac{u}{v}) = \frac{V\frac{du}{dx}-U\frac{dv}{dx}}{V^2}[/tex]

letting [tex]u = E^2R[/tex] and [tex]v = (R+r)^2[/tex]

So ...

[tex]\frac{dP}{dR}=\frac{E^2(R+r)^2 - 2(R+r)(E^2R)}{(R+r)^{4}}[/tex]

at maximum power [tex]\frac{dP}{dR}=0[/tex]

[tex](R+r)^{4}[/tex] can not equal zero, therefore [tex]E^2(R+r)^2 - 2(R+r)(E^2R)[/tex] must equal zero.

Rearranging this eventually gets to r = R, thus showing at maximum power the load resistance equals the internal resistance

The R resistance, also known as the external resistance, is equal to the internal resistance because they both contribute to the overall resistance in a circuit. The external resistance represents the resistance of the circuit components, such as wires and resistors, while the internal resistance represents the resistance of the source of electricity, such as a battery. These two resistances work together to limit the flow of current in a circuit, and when added together, they equal the total resistance.

The internal resistance of a source of electricity, such as a battery, adds to the overall resistance of a circuit. This means that the higher the internal resistance, the higher the total resistance will be. This is because the internal resistance acts as a limiting factor for the flow of current, making it more difficult for electricity to flow through the circuit.

No, the internal resistance cannot be ignored in circuit calculations. It plays an important role in determining the overall resistance and current in a circuit. Ignoring the internal resistance can lead to inaccurate calculations and may affect the functioning of the circuit.

The internal resistance of a source of electricity can be measured by connecting a known external resistance to the source and measuring the voltage across the external resistance. By using Ohm's Law (V = IR), the current can be calculated, and the internal resistance can be determined using the formula: r = (E - V)/I, where r is the internal resistance, E is the emf of the source, V is the voltage across the external resistance, and I is the current.

The internal resistance of a source of electricity can be affected by various factors such as the type of material used in the battery, the temperature of the battery, and the age and condition of the battery. Generally, as a battery ages, its internal resistance increases, leading to a decrease in its ability to supply current to a circuit.

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