Question

an electric circuit contains an 8-ohm resistor in series with an inductor of 0.5 Henry and a battery of E volts. at t= 0, the current is 0. find E if the maximum current flowing in the circuit is 8 amperes

Answer 1

in this question the resistor and the inductor are in a series connection,
so at a time t->0('t' tending to zero(a limit)), the current in the circuit is zero.
and at t=infinity the inductor looses its battle with the initial potential difference E and offers no back EMF(or impedance). so there is no longer any role of the inductor ,then E/R will be our max. current.
we have E/R=8 and,\[R=8 \Omega\]so E=8x8=64V

Answer 2

@SATYA.A.TEJA can you do this using differential equations? step-by-step?

Answer 3

E=i(t)R+Ldi(t)/dt
Solving the homogeneous equation: di(t)/i(t)=-(R/L)dt
Then Log[i(t)]=-(R/L)t + K1 or i(t)=A1+A2*exp[-(R/L)t]

Now apply boundary condition at t=0 i(0)=0----->A1+A2=0 and i(t)=A1*(1-exp[-(R/L)t])
Replacing i(t) in the differential equation, it turns out to be A1=E/R

Then i(t)=E/R*(1-exp[-(R/L)t])

The maximum current happens when t->infinite and is E/R, then E=R*i(max)=8x8=64V

Answer 4

is this the same as that equation \[\frac{dI}{dt} + (\frac{L}{R})I = \frac{E}{R}\]

Answer 5

No, it is not the same. In fact it is: dI/dt+(R/L)I=E/L

Answer 6

do you know how to use the formula i stated?

Answer 7

The formula you stated is incorrect:
Voltage in the coil = V1=LdI/dt
Voltage in the resistor=V2=RI
Voltage in the Battery=E
It has to comply with Kirchoff's Law: E=V1+V2===>E=LdI/dt+RI (my formula, solved in my first answer)

Answer 8

but this is what my teacher gave me...

Answer 9

okay..can you just explain your formula step by step?

Answer 10

just rechecked...it is R/L sorry so how to do this step by step?

Answer 11

Here it goes. Enjoy it!

Answer 12

thanks! will take a look