Certainly! Let's solve this step-by-step.
### Given Data:
- Electromotive force (emf) of the cell, [tex]\( \text{emf} = 2 \)[/tex] volts
- Internal resistance of the cell, [tex]\( r = 1 \ \Omega \)[/tex]
- Current supplied by the cell, [tex]\( I = 0.5 \)[/tex] amperes
### Goal:
Find the resistance of the external circuit, which we'll denote as [tex]\( R \)[/tex].
### Ohm's Law and Total Voltage:
According to Ohm's Law, the relationship between voltage ([tex]\( V \)[/tex]), current ([tex]\( I \)[/tex]), and resistance ([tex]\( R \)[/tex]) is given by:
[tex]\[ V = I \cdot R \][/tex]
The total voltage provided by the cell (emf) is the sum of the voltage drops across the internal resistance and the external resistance. So, we can write:
[tex]\[ \text{emf} = I \cdot (R + r) \][/tex]
### Rearranging to Solve for [tex]\( R \)[/tex]:
We need to isolate [tex]\( R \)[/tex] in the equation:
[tex]\[ \text{emf} = I \cdot (R + r) \][/tex]
Thus:
[tex]\[ R + r = \frac{\text{emf}}{I} \][/tex]
Subtract the internal resistance [tex]\( r \)[/tex] from both sides to solve for [tex]\( R \)[/tex]:
[tex]\[ R = \frac{\text{emf}}{I} - r \][/tex]
### Plugging in the Given Values:
[tex]\[ R = \frac{2 \text{ volts}}{0.5 \text{ amperes}} - 1 \ \Omega \][/tex]
Calculate the division:
[tex]\[ \frac{2}{0.5} = 4 \][/tex]
Subtract the internal resistance:
[tex]\[ R = 4 - 1 \][/tex]
Therefore:
[tex]\[ R = 3 \ \Omega \][/tex]
### Conclusion:
The resistance of the external circuit is [tex]\( 3 \ \Omega \)[/tex].
