Answer:
R = 2481 Ω
L= 1.67 H
Explanation:
(a) We have an inductor L which has an internal resistance of R. The inductor is connected to a battery with an emf of E = 12.0 V. So this circuit is equivalent to a simple RL circuit. It is given that the current is 4.86 mA at 0.725 ms after the connection is completed and is 6.45 mA after a long time. First we need to find the resistance of the inductor. The current flowing in an RL circuit is given by
i = E/R(1 -e^(-R/L)*t) (1)
at t --> ∞ the current is the maximum, that is,
i_max = E/R
solve for R and substitute to get,
R= E/i_max
R = 2481 Ω
(b) To find the inductance we will use i(t = 0.940 ms) = 4.86 mA, solve (1) for L as,
Rt/L = - In (1 - i/i_max )
Or,
L = - Rt/In (1 - i/i_max )
substitute with the givens to get,
L = -(2481 Si) (9.40 x 10-4 s)/ In (1 - 4.86/6.45 )
L= 1.67 H
note :
error maybe in calculation but method is correct
(a) 1860 Ω
(b) 0.16H
This is an RL circuit as it contains an inductor with and inherent resistance.
The current, I, flowing through such circuit is given by;
I = [tex]\frac{E}{R}[/tex][1 - [tex]e^{-(R/L)t}[/tex]] --------------(i)
Where;
E = emf of the circuit = 12.0V
R = Resistance of the inductor
L = inductance of the inductor
t = time taken for the flow of current
(a) After a long time, i.e t → ∞, the current becomes saturated (i.e maximum) and the value is 6.45mA.
At this time, [tex]e^{-(R/L)t}[/tex] (from equation (i)) becomes 0 and current I, becomes;
I = [tex]\frac{E}{R}[/tex] -------------------(ii)
Now substitute the value of E and I into equation (ii) as follows;
6.45 x 10⁻³= [tex]\frac{12.0}{R}[/tex]
R = [tex]\frac{12.0}{6.45 * 10^{-3}}[/tex]
R = 1860Ω
Therefore, the resistance of the inductor is 1860 Ω
(b) To get the inductance L of the inductor, we substitute I = 4.86mA at time t = 0.725ms into equation (i) as follows;
4.86 x 10⁻³ = [tex]\frac{12.0}{1860}[/tex][1 - [tex]e^{-(1860/L)0.00486}[/tex]]
4.86 x 10⁻³ x 1860 = 12.0[1 - [tex]e^{-(1860/L)0.00486}[/tex]]
0.7533 = [1 - [tex]e^{-(1860/L)0.00486}[/tex]]
1 - 0.7533 = [tex]e^{-(1860/L)0.00486}[/tex]
0.2467 = [tex]e^{-(1860/L)0.00486}[/tex]
Take the natural log of both sides
ln(0.2467) = -1860 x 0.00468 / L
-1.3996 = -8.7048 / L
L = -1.3996 / -8.7048
L = 0.16H
Therefore, the inductance of the inductor is 0.16H
