Neither do I, but assuming it is...BryceHarper said:The current in a series RL circuit increases to 20% of its final value in 3.4μs. If L = 2.0mH, what is the resistance R?
I=I_0(1-e^(-t/(L/R))) <----Im not sure if this equation is right?
A series RL circuit is an electrical circuit that consists of a resistor (R) and an inductor (L) connected in series with a power source. The current in this type of circuit flows through both components in a single loop.
The inductor in a series RL circuit stores energy in the form of a magnetic field. As the current flows through the inductor, the magnetic field builds up and resists any changes in the current flow. This property of inductors is what causes a delay in the current in a series RL circuit.
In a series RL circuit, the current will initially be low due to the inductor's resistance to changes in current. As the inductor's magnetic field builds up, the current will gradually increase until it reaches a steady state. At this point, the current will be determined by the resistance of the circuit and the voltage of the power source.
The equation for calculating the current in a series RL circuit is I = V/R, where I is the current, V is the voltage of the power source, and R is the total resistance of the circuit. This equation assumes that the inductor has reached a steady state and the current is constant.
If the resistance in a series RL circuit is increased, the current will decrease as the overall resistance in the circuit increases. On the other hand, if the inductance is increased, the current will decrease initially due to the inductor's resistance, but then it will gradually increase to reach a steady state with a higher current value.