Answer:
t = 0.69R/L
If the resistance R increases, the required time decreases, for the battery reaches a 50% of its initial value.
Explanation:
In order to know what happens to the the time, when the battery reaches a 50% of its initial voltage, while the RL resistance increases, you use the following formula:
[tex]V=V_oe^{-\frac{R}{L}t}[/tex] (1)
Vo: initial voltage in the battery
V: final voltage in the battery = 0.5Vo
R: resistance of the RL circuit
L: inductance of the RL circuit
You use properties of logarithms to solve the equation (1) for t:
[tex]0.5V_o=V_oe^{-\frac{R}{L}t}\\\\ln(0.5)=-\frac{R}{L}t\\\\t=-\frac{L}{R}ln(0.5)=0.69\frac{L}{R}[/tex] (2)
By the result obtained in the equation (2), you can observe that if the resistance R increases, the required time decreases, for the battery reaches a 50% of its initial value.
