Answer to Question #193282 in Math for Allysa

Question

An inductance of L Henrys and a resistance of 8 ohms are connected in 

series with EMF of 100 Volts. If the current is initially zero and is equal to 18

amperes after 1 second. Find L and find the current after 0.2 s.

Answer 1

Applying Kirchhoff's loop rule to this circuit gives

$$E_o - IR - L \frac{dI}{dt}=0$$

The equation can be solved by separating the variables and integrating

$$\intop \frac{R}{L} dt=\int \frac{dI}{E_o/R -I}$$

The result is

$$I(t)=\frac {E_o}{R} \Big (1-e^{- \frac {R}{L}t} \Big)$$

The final value of the current can be obtained by setting dI/dt equal to zero:

$$I_f = \frac {E_o}{R}=\frac {100V}{8 \Omega}=12.5 A$$

In this circuit, no current flow of 18A is possible. To obtain a current of 18A, it is necessary either to decrease the resistance value or to increase the EMF.

For example, consider a solution for double EMF. Then

$$I_f = \frac {E_o}{R}=\frac {200V}{8 \Omega}=25 A$$

So, after 1 second

$$18A=\frac{200V}{8 \Omega} \Big (1-e^{- \frac {8 \Omega}{L}1s} \Big)$$

then

$$L=\frac {-8 \Omega}{ln(1-\frac{18A \cdot 8\Omega}{200V})}\approxeq 6.3H$$

the current after 0.2s

$$I(0.2)=\frac {200V}{8\Omega} \Big (1-e^{- \frac {8\Omega}{6.3H}0.2s} \Big)\approxeq 5.6A$$

Another example, consider a solution for a resistance half the specified value. Then

$$I_f = \frac {E_o}{R}=\frac {100V}{4 \Omega}=25 A$$

So, after 1 second

$$18A=\frac{100V}{4 \Omega} \Big (1-e^{- \frac {4 \Omega}{L}1s} \Big)$$

then

$$L=\frac {-4 \Omega}{ln(1-\frac{18A \cdot 4\Omega}{100V})}\approxeq 3.1H$$

the current after 0.2s

$$I(0.2)=\frac {100V}{4\Omega} \Big (1-e^{- \frac {4\Omega}{3.1H}0.2s} \Big)\approxeq 5.7A$$