To solve this problem, we can use the concept of power in an electric circuit. The power delivered to a resistor can be calculated using the equation P = I^2 * R, where P is the power, I is the current, and R is the resistance.
Let's start by calculating the current in the circuit when the battery is connected to the 100 Ω resistor. We can use the formula for power to find the current:
1.588 J/s = I^2 * 100 Ω
I^2 = 1.588 J/s / 100 Ω
I^2 = 0.01588 A^2
I = sqrt(0.01588 A^2)
I ≈ 0.126 A
Now, let's calculate the current when the battery is connected to the 200 Ω resistor:
0.802 J/s = I^2 * 200 Ω
I^2 = 0.802 J/s / 200 Ω
I^2 = 0.00401 A^2
I = sqrt(0.00401 A^2)
I ≈ 0.063 A
Now that we have the current values, we can calculate the emf (E) and the internal resistance (r) of the battery.
Using the equation for the emf of the battery, E = I * (R + r), we can substitute the values we have:
For the first case:
E = 0.126 A * (100 Ω + r)
For the second case:
E = 0.063 A * (200 Ω + r)
Now we have a system of two equations with two variables (E and r). We can solve this system to find the values of E and r.
First, let's isolate r in the first equation:
E = 0.126 A * (100 Ω + r)
E = 12.6 ΩA + 0.126 A * r
r = (E - 12.6 ΩA) / 0.126 A
Now, substitute this expression for r in the second equation:
E = 0.063 A * (200 Ω + (E - 12.6 ΩA) / 0.126 A)
Simplifying this equation will give us the value of E, the emf of the battery. Once we have E, we can substitute it back into the expression for r to find the value of r, the internal resistance of the battery.
I hope this helps you to solve the problem! If you have any further questions, please feel free to ask.