Rate of current changing through an inductor

[supersunshine]

Homework Statement

At time t=0 a 45V potential difference is applied to the leads of a coil with inductance L=50 mH and resistance R=180 ohm. At what rate is the current through the coil increasing at t=1.2ms?

Homework Equations

1) L=EMF/ (di/dt)
2) L=N(flux)/i
3)emf=iR

The Attempt at a Solution

I thougt you could first find the current at t=o using EMF=iR= .25 A, then you could find the current at t=1.2 ms, by first finding the flux at this time to do this:

I thought -d(flux)/dt=EMF= 45V. Is it correct to say that the magnetic flux is changing at a rate of 45 wb/s ? and then multiply by time and use equation 2 , however we don't know the number of turns, which is where I am stuck

any hints would be appreciated

 

[Astronuc]

This is a simple RL circuit.

http://farside.ph.utexas.edu/teaching/302l/lectures/node88.html

 

[m2003]

Firstly, it is important to note that the units for inductance (L) are in henrys (H), not webers (wb).

To solve this problem, we can use the equation L=EMF/(di/dt), where L is the inductance, EMF is the electromotive force (voltage), and (di/dt) is the rate of change of current.

At t=0, the EMF is 45V and the current is 0A. So we can rearrange the equation to solve for (di/dt) at t=0: (di/dt)=EMF/L= (45V)/(50mH)= 900 A/s.

To find the rate of change of current at t=1.2ms, we can use the same equation, but with a different value for EMF. At t=1.2ms, the EMF is still 45V, but the current is no longer 0A. We can use Ohm's Law (V=IR) to find the current at t=1.2ms: I=(45V)/(180 ohm)= 0.25 A.

Plugging these values into the equation (di/dt)=EMF/L, we get: (di/dt)=(45V)/(50mH)= 900 A/s. This means that the current is increasing at a rate of 900 A/s at t=1.2ms.