Answer:
[tex]36 kJ; 9\times10^5s = 25h[/tex]
Explanation:
Let's assume the circuit runs DC with a purely resistive load, and every component has efficiency 1, so no energy is dissipated.
The power generated by the cell is equal to the power absorbed by the load, which is the product of the tension drop (equal to the emf generated by the cell by KVL) times the current flowing through it: [tex]2.0V\times 0.5A = 1.0 W[/tex]. Remembering the definition of power and the relationship with energy, we obtain
[tex]P=\frac Et \rightarrow E = 1W\times 10h = 1W\times 36000 s = 36 kJ[/tex]
If the current becomes [tex]0.2A[/tex], and the energy stored inside the cell doesn't change, you just scale up by a factor [tex]0.5/0.2= 2.5[/tex], thus taking 25 hours to discharge. How?
The power absorbed with the new, smaller current, is [tex]2.0V \times 0.2A= 0.4W[/tex]. Back to the definition of power, replacing this new value and keeping the same energy value in the cell we get
[tex]P= \frac Et \rightarrow t= \frac E P = \frac {3.6\times 10^4J}{0.4W}= 9\times 10^5s = 25h[/tex]
