Question

A battery of emf E and internal resistance r sends a current, I1 , I2 when connected to an external resistance of R1 , R2 respectively. Find the emf. and internal resistance of the battery.

Answer 1

Let's derive the expressions for the electromotive force (EMF) and the internal resistance of the battery step-by-step.

Understanding the Setup

The core principle we'll use is Ohm's law applied to the entire circuit. For a battery with EMF ($E$) and internal resistance ($r$) connected to an external resistance ($R$), the total resistance in the circuit is $(R + r)$.

The current ($I$) flowing through the circuit is given by:
$I = \frac{\text{Total EMF}}{\text{Total Resistance}}$
$I = \frac{E}{R + r}$

This can be rearranged to a more useful form:
$E = I(R + r)$

Setting up the Equations

We are given two different scenarios, which we can use to create a system of two equations with two unknowns ($E$ and $r$).

Scenario 1:
External Resistance = $R_1$
Current = $I_1$

Using our formula, we get our first equation:
$E = I_1(R_1 + r)$
$E = I_1R_1 + I_1r \quad \cdots \text{(Equation 1)}$

Scenario 2:
External Resistance = $R_2$
Current = $I_2$

Using our formula, we get our second equation:
$E = I_2(R_2 + r)$
$E = I_2R_2 + I_2r \quad \cdots \text{(Equation 2)}$

Solving for Internal Resistance (r)

Since the EMF ($E$) is the same in both equations, we can set Equation 1 and Equation 2 equal to each other to eliminate $E$.

$I_1(R_1 + r) = I_2(R_2 + r)$

Now, we solve for $r$:
1. Expand the terms:

$I_1R_1 + I_1r = I_2R_2 + I_2r$

1. Group all terms with $r$ on one side and all other terms on the other side:
   $I_1r - I_2r = I_2R_2 - I_1R_1$

2. Factor out $r$ from the left side:
   $r(I_1 - I_2) = I_2R_2 - I_1R_1$

3. Isolate $r$ by dividing:
   $r = \frac{I_2R_2 - I_1R_1}{I_1 - I_2}$

To make the expression look a bit neater, we can multiply the numerator and the denominator by -1:
$r = \frac{-(I_2R_2 - I_1R_1)}{-(I_1 - I_2)} = \frac{I_1R_1 - I_2R_2}{I_2 - I_1}$

So, the internal resistance is:
$$ r = \frac{I_1R_1 - I_2R_2}{I_2 - I_1} $$

Solving for EMF (E)

Now that we have an expression for $r$, we can substitute it back into either Equation 1 or Equation 2 to find $E$. Let's use Equation 1:

$E = I_1(R_1 + r)$

Substitute the expression for $r$:
$E = I_1 \left( R_1 + \frac{I_1R_1 - I_2R_2}{I_2 - I_1} \right)$

To simplify, find a common denominator inside the parenthesis:
$E = I_1 \left( \frac{R_1(I_2 - I_1) + (I_1R_1 - I_2R_2)}{I_2 - I_1} \right)$

Expand the terms in the numerator:
$E = I_1 \left( \frac{R_1I_2 - R_1I_1 + I_1R_1 - I_2R_2}{I_2 - I_1} \right)$

The terms $-R_1I_1$ and $+I_1R_1$ cancel out:
$E = I_1 \left( \frac{R_1I_2 - I_2R_2}{I_2 - I_1} \right)$

Factor out $I_2$ from the numerator:
$E = I_1 \left( \frac{I_2(R_1 - R_2)}{I_2 - I_1} \right)$

Combine the terms to get the final expression for EMF:
$$ E = \frac{I_1 I_2 (R_1 - R_2)}{I_2 - I_1} $$

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Summary of Results

The internal resistance ($r$) and the EMF ($E$) of the battery are given by the following expressions:

Internal Resistance (r):
$$ r = \frac{I_1R_1 - I_2R_2}{I_2 - I_1} $$

EMF (E):
$$ E = \frac{I_1 I_2 (R_1 - R_2)}{I_2 - I_1} $$