A circuit consists of a self inductor of L=0.003 H in series with a resistor R1=5 Ohm. Parallel to these is a resistor R2=10 Ohm. A battery of V0=9 volt is driving the circuit.
Current has been running for 10 minutes.
(a) How much energy (in Joules) is now stored in the self-inductor.
(b) How much power (in Watt) is then generated by the battery into the circuit? (ignore internal resistance of the battery).
The connection to the battery is now broken (so that the battery is not connected to the circuit anymore).
(c) How long will it take (in seconds) for the current through R1 to be reduced by 50%?
(d) How long does it take (in seconds) till the energy stored in the self-inductor has been reduced by 50%?
(e) How long will it take (in seconds) for the current in R2 to be reduced by 50% (compared to the highest value right after the battery is disconnected)?
plz help.....!
I can tell you a) & b)
a) Assuming after 10 min I=Imax then
I1=V/R1 and so the energy is 1/2*L*I1
b) I=V/Req (Req=(R1*R2)/(R1+R2)) So P=I*V
Thnx :) a) n b) are correct ! plz some one for c), d) n e) parts ???
For c) & e):For broken connection of V then Req is R1+R2 so the equation is :
Imax/2=Imax*e^((-Req/L)*t) and Imax is the Imax from before the connection with the V was broken so Imax=V/R where R=(R1*R2)/(R1+R2)
I still can't get the solution for d if i manage to get it i'll post it
can u plz tell the answers in numerals for time in parts c) n e). I am unable to solve it correctly !
0.0001386 both. Spent a little time and put the equation in Wolfram alpha it's online calculator and it's very helpfull in general
thnx c n e are correct !
plz lemme knw the b) part when u are able to figure out !
d)since you have find a) then first you should find what the current is for the half energy 0.00486/2=0.00243 so 0.00243=1/2*0.003*I^2 so I=1.27279.the max current in the self inductor was i=9/5=1.8A so the equation is 1.27279=1.8*e^(-(15/0.003)*t)(R=(5+10)Ù because now the connection with v is broken)so t=0.0000693135
Can somebody help with Problem: Displacement Current please?
I thought the formula is:
(1.25663706*10^-6*0.04*0.2*0.0705)/(2*pi*0.05^2)
However, I am not getting the correctr answer.
FLu the formoula is (ì0*I)/(2*pi*r) and ì0=4*pi*10^-7
Thanks Anonymous!
Do you have by any chance the answer for Problem:
Opening a Switch on an RL Circuit
b) and c)
I could figure out a)V_0/R_1
the other one seems more complicated.
Can you help please?
FLu b) is the equation
I0*e^((-(R1+R2)/L)*t) I had no luck with the others
Many thanks Anonymous!
I got d) 100
Opening Switch on an RL Circuit:
b) seems in a way easier as most of the letters are not accepted only R1, R2 and V0.
Did somebody managed it yet?
a and d please
dd, I have provided a and b already just have a look above.
Anyone for c) please?
Did somebody found c) yet please!?
Sorry meant a and d. b) was provided by Anonymous.
If anybody knows c), thanks in advance!
Yes, please answer c) please!
c) please!
c) Anyone?
Anyone for c) guys please?
Was someone successful with c)?
for C)... divide the b) by 2
PLEASE NUMERICAL ANSWERS FOR A) AND B)
Thanks Supraconductor.
I could not figure it out. Is it this:
V_0*e^(R_1+R_2)
Thanks in advance!
Having problems with c) is well, was somebody able to come up with the expression?
Expression for c) please?
I tried it out now with the hint of Supraconductor but could not figured out. Did somebody manage it please?
c) please anyone!
Please c)?
c)0.0001386
Thanks hmmmm, I meant Problem:
Opening a Switch on an RL Circuit
Do you have c)?
Anyone for Problem:
Opening a Switch on an RL Circuit: c)?
Thanks.
Finally c)(V_0*(R_1+R_2))/R_1
(Part c for RL circuit)
Great many thanks Anonymous!
Please tell RL Circuit I2 I3 for b and c
ASAP