I am given:
Kirchhoff’s voltage law states that the sum of the voltage drops across an inductor, L dI/dt, and across a resistor, IR, must be the same as the voltage source, E(t), applied to the circuit. The resulting ODE model is LdI/dt+IR=E(t). If a 12-volt battery is connected to an RL circuit with a ½ henry inductor, L, and a 10 ohm resistor, R , find the current I(t) given that I(0)=0.
Ok, so I understand the formula is $E(t)=L(dI/dt)+IR$, but what is confusing me is with the mention of I(t). How do I begin to solve this?
Thanks!
Best Answer
Hint: this is a separable equation:
$$L(dI/dt) = E - IR$$
So, we can separate and integrate:
$$\int \dfrac{L}{E-IR}~dI = \int dt$$