Asolenoid having an inductance of 6.95 μh is connected in series with a 1.24 kω resistor. (a) if a 12.0 v battery is connected across the pair, how long will it take in seconds for the current through the resistor to reach 73.6% of its final value? (b) what is the current through the resistor at a time t = 1.00τl?

In electrical circuit, this arrangement is called a R-L series circuit. It is a circuit containing elements of an inductor (L) and a resistor (R). Inductance is expressed in units of Henry while resistance is expressed in units of ohms. The relationship between these values is called the impedance, denoted as Z. Its equation is

Z = √(R^2 + L^2)
Z =  √((1.24×10^3 ohms)^2 + (6.95×10^-6 H)^2)
Z = 1,240 ohms

The unit for impedance is also ohms. Since the circuit is in series, the voltage across the inductor and the resistor are additive which is equal to 12 V. Knowing the impedance and the voltage, we can determine the maximum current.
I = V/Z=12/1,240 = 9.68 mA
But since we only want to reach 73.6% of its value, I = 9.68*0.736 = 7.12 mA. Then, the equation for R-L circuits is
, where τ = L/R = 6.95×10^-6/1.24×10^3 = 5.6 x 10^-9
Then,


t = 7.45 nanoseconds
Part B.) If t = 1.00τ, then t/τ = 1. Therefore,

 
I = 6.12 mA 

Where are the options ?

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