College physics homework question, see figure below: There is a 12 volt battery,
ID: 1769485 • Letter: C
Question
College physics homework question, see figure below:
There is a 12 volt battery, which is connected to the two resistors (R1 is 5 omega and R2 is 7 omega) and an inductor (L is 3 mH). Switch S has been open for a long time and is closed at time t=0 s.
1. Compute the magnitude of the current I(L) immediately after the switch is closed.
2. Compute the magnitude of dI(L)/dt immediately after the switch is closed.
3. What is the magnitude of the current I(L) after the switch has been closed for a long time?
4. After the switch has been closed for a long time, it is opened. The current I(L) now decays exponentially as a function of time. What is the value of the time constant of this decay?
All the answers will be thankful. The correct and clear answer will be rated as five stars and 1500 points. Thanks!
Explanation / Answer
a)
Immediately afer switch is closed ,inductor acts as open circuit ,so current through inductor is
IL=0 A
b)
Total Current flowing in the circuit
I=V/(R1+R2) =12/(5+7) =1 A
Voltage across Inductor
VL=V2=I*R1=1*7
VL=7 Volts
since
VL=LdIL/dt
=>dIL/dt =VL/L =7/(3*10-3) =2333.33 A/s
c)
After Long time indcutor acts as short circuit ,so current flowing through inductor is
IL=V/R1=12/5 =2.4 A
d)
New time Constant
T=L/R2=3*10-3/7
T=4.2857*10-4 seconds or 0.42857 ms