An inductor of inductance L= 400mH and resistors R1=2Ω and R2=2Ω are connected to a battery of emf 12V . The internal resistance of the battery is negligible. The switch S is closed at t=0. The potential drop across L as a function of time is:

 oltage across the inductor in the circuit is proportional to the rate of current flow through the inductor. For this first calculate the current flowing through the branch containing L and R2. Then differentiate current i with respect to time t.

Given that,




E= 12V

In the branch containing L and R2, i=E/R1−e−R2t/L

Differentiating current, i with respect to time t we get,

⇒ di/d= ER2e−R2t/L. R2/L = E/Le−R2t/L

Therefore, VL = L di/d=Ee−R2t/L = 12e−5tV

Then the potential drop across L as a function of time is 12e−5tV.

Getting Info...
Cookie Consent
We serve cookies on this site to analyze traffic, remember your preferences, and optimize your experience.
It seems there is something wrong with your internet connection. Please connect to the internet and start browsing again.
AdBlock Detected!
We have detected that you are using adblocking plugin in your browser.
The revenue we earn by the advertisements is used to manage this website, we request you to whitelist our website in your adblocking plugin.
Site is Blocked
Sorry! This site is not available in your country.