To solve this problem, we will use Ohm's Law and the power formula.
First, let's analyze the first scenario where the battery is connected to a 100 Ω resistor.
We are given that the battery delivers 1.588 J/s to the 100 Ω resistor. This is the power being supplied by the battery. We can use the power formula to find the current flowing through the circuit.
Power = Current^2 * Resistance
1.588 J/s = (Current)^2 * 100 Ω
Solving for Current:
Current = sqrt(1.588 J/s / 100 Ω)
Current ≈ 0.398 A
Now, let's analyze the second scenario where the battery is connected to a 200 Ω resistor.
We are given that the battery delivers 0.802 J/s to the 200 Ω resistor. Again, this is the power being supplied by the battery. We can use the power formula to find the current flowing through the circuit.
Power = Current^2 * Resistance
0.802 J/s = (Current)^2 * 200 Ω
Solving for Current:
Current = sqrt(0.802 J/s / 200 Ω)
Current ≈ 0.2 A
Now that we have the currents in both scenarios, we can use Ohm's Law to find the emf (ε) and internal resistance (r) of the battery.
In the first scenario:
ε - (Current) * r = (Current) * 100 Ω
ε - 0.398 A * r = 0.398 A * 100 Ω
In the second scenario:
ε - (Current) * r = (Current) * 200 Ω
ε - 0.2 A * r = 0.2 A * 200 Ω
We now have a system of equations with two unknowns (ε and r). Solving this system of equations will give us the values of the emf and internal resistance.
To draw suitable diagrams, you can represent the battery as a symbol with the emf (ε) and internal resistance (r) labeled. Connect the battery to the resistors using wires, and label the resistors with their respective values (100 Ω and 200 Ω). You can also label the currents flowing through the circuit.