A 22 mH inductor, an 14.0 resistor, and a 6.0 V battery are connected in series. The switch is closed at t 0.Find the voltage drop across the resistor at the following times.(a) t 0V(b) after one time constant has passedVAlso, find the voltage drop across the inductor at the following times.(c) t 0V(d) after one time constant has elapsedV
The final inductor current will be .428 and the voltage will be 0. A) 0 B) .632*.428.271 C) 14 D)14*(1-.632)5.15
oubaas: Can you explain why for the inductor you don't use the 1-e^whatever and for the resistor you do? I'm just trying to understand the concepts before for my midterm. Thanks!
? R*C 22*14 0.308 sec At t 0 - voltage across resistor Vr V(1-e^-0) V*0 0 - voltage across inductor Vl V(e^-0) V*1.00 6V At t ? - voltage across resistor Vr V(1-e^-1) V*0.632 3.792 V - voltage across inductor Vl V(e^-1) V*0.368 2.208 V
B is the respond because of the fact: At t 0, the inductor is modeled as an open circuit because of the fact an inductor grants very severe impedance (reactance) for as we communicate changing present day. the final public of the voltage interior the ckt authentic after the swap is closed looks around the inductor. we are able to write the equation for the present interior the ckt as I(t) [20/2][a million-e^-(t*R/L)] observe that at t0, I(0) 10*[a million- e^(0)] 10*[a million-a million] 0 For t very great the term e^-(t*R/L) approaches 0 That leaves I(t) 10A. the present interior the inductor and resistor are a similar because of the fact they are in sequence.
