The first half of my answer is just to derive a result which you can then choose to memorize for future reference.
Applying Kirchhoff's loop rule stating that the circuit's voltage gains/drops sum to 0:
ΣV = 0
When we have a direct current resistor-inductor circuit where a battery, resistor, and inductor are connected in series, we have the following voltage values for each component as listed below. Positive values indicate a voltage gain and negative values indicate a voltage drop:
Sum up these voltage changes and set the sum equal to 0 as per Kirchhoff's loop rule:
ℰ - iR - L(di/dt) = 0
What we have here is a first-order differential equation because we have the above equation in terms of current as a function of time as well as its first derivative with respect to time. What we want to do is find a current function that both satisfies the differential equation as well as the initial condition that the current at t = 0s is zero, because we have just inserted the battery and are waiting for the current through the circuit to build up.
Actually finding such a current function requires you to have taken at least an introductory class on differential equations... or access to the Wolfram|Alpha differential equation solver. Since the rest of the derivation is all mathematics, I will take you straight to the result.
The current function i(t) that satisfies both the differential equation above as well as the initial condition that i(0) = 0 is given by:
i(t) = (-ℰ/R)[tex]e^{-Rt/L}[/tex] + ℰ/R
We know that the inductor will oppose the change in current, but as time goes on this opposition decreases closer and closer to 0. To find the maximum current, we simply calculate:
[tex]\lim_{t \to \infty} i(t)[/tex] = ℰ/R
The maximum current is simply the battery's terminal voltage divided by the resistor's resistance.
Now we calculate when the circuit reaches 50.0% of the maximum current, i.e. the time t for which i = ℰ/(2R)
This is easy algebra from here on to the end:
i(t) = (-ℰ/R)[tex]e^{\frac{-Rt}{L}}[/tex]+ℰ/R = ℰ/(2R)
[tex]-e^{\frac{-Rt}{L}}[/tex] + 1 = 1/2
[tex]e^{\frac{-Rt}{L}}[/tex] = 1/2
-Rt/L = ㏑(1/2)
t = (-L/R)㏑(1/2)
Substitute our given values for L and R:
L = 3.50H, R = 12.0Ω
t = 0.202s
Therefore, the circuit will reach half of its maximum current at 0.202 seconds.
The time interval will the current reach 50.0% of its final value is 0.2 s.
The rate at which current in the ciruit drops is given by the following formulas;
i(t) = (-ℰ/R)+ℰ/R = ℰ/(2R)
[tex]-e^{\frac{-Rt}{L} } + 1= \frac{1}{2} \\\\e^{\frac{-Rt}{L} } = \frac{1}{2}[/tex]
where;
-Rt/L = ㏑(1/2)
t = (-L/R)㏑(1/2)
t = (-3.5/12) ln(0.5)
t = 0.2 s
Thus, the time interval will the current reach 50.0% of its final value is 0.2 s.
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