Asked by abdul rahman
                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)?
            
            
        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)?
Answers
                    Answered by
            abdul rahman
            
    plz help.....!
    
                    Answered by
            s
            
    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
    
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
                    Answered by
            abdul rahman
            
    Thnx :) a) n b) are correct ! plz some one for c), d) n e) parts ???
    
                    Answered by
            s
            
    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)
    
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)
                    Answered by
            s
            
    I still can't get the solution for d if i manage to get it i'll post it
    
                    Answered by
            abdul rahman
            
    can u plz tell the answers in numerals for time in parts c) n e). I am unable to solve it correctly !
    
                    Answered by
            s
            
    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
    
                    Answered by
            abdul rahman
            
    thnx c n e are correct !
    
                    Answered by
            abdul rahman
            
    plz lemme knw the b) part when u are able to figure out !
    
                    Answered by
            Anonymous
            
    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
    
                    Answered by
            FLu
            
    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.
    
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.
                    Answered by
            Anonymous
            
    FLu the formoula is (ì0*I)/(2*pi*r) and ì0=4*pi*10^-7
    
                    Answered by
            FLu
            
    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?
    
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?
                    Answered by
            Anonymous
            
    FLu b) is the equation 
I0*e^((-(R1+R2)/L)*t) I had no luck with the others
    
I0*e^((-(R1+R2)/L)*t) I had no luck with the others
                    Answered by
            FLu
            
    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?
    
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?
                    Answered by
            dd
            
    a and d please
    
                    Answered by
            FLu
            
    dd, I have provided a and b already just have a look above.
Anyone for c) please?
    
Anyone for c) please?
                    Answered by
            My
            
    Did somebody found c) yet please!?
    
                    Answered by
            FLu
            
    Sorry meant a and d. b) was provided by Anonymous.
If anybody knows c), thanks in advance!
    
If anybody knows c), thanks in advance!
                    Answered by
            Ur
            
    Yes, please answer c) please!
    
                    Answered by
            Lora
            
    c) please!
    
                    Answered by
            Gema
            
    c) Anyone?
    
                    Answered by
            FLu
            
    Anyone for c) guys please?
    
                    Answered by
            My
            
    Was someone successful with c)?
    
                    Answered by
            Supraconductor
            
    for C)... divide the b) by 2
    
                    Answered by
            Anony
            
    PLEASE NUMERICAL ANSWERS FOR A) AND B)
    
                    Answered by
            FLu
            
    Thanks Supraconductor.
I could not figure it out. Is it this:
V_0*e^(R_1+R_2)
Thanks in advance!
    
I could not figure it out. Is it this:
V_0*e^(R_1+R_2)
Thanks in advance!
                    Answered by
            My
            
    Having problems with c) is well, was somebody able to come up with the expression?
    
                    Answered by
            Haet
            
    Expression for c) please?
    
                    Answered by
            Ula
            
    I tried it out now with the hint of Supraconductor but could not figured out. Did somebody manage it please?
    
                    Answered by
            Terry
            
    c) please anyone!
    
                    Answered by
            Ur
            
    Please c)?
    
                    Answered by
            hmmmm
            
    c)0.0001386
    
                    Answered by
            Ur
            
    Thanks hmmmm, I meant Problem:
Opening a Switch on an RL Circuit
Do you have c)?
    
Opening a Switch on an RL Circuit
Do you have c)?
                    Answered by
            FLu
            
    Anyone for Problem:
Opening a Switch on an RL Circuit: c)?
Thanks.
    
Opening a Switch on an RL Circuit: c)?
Thanks.
                    Answered by
            Anonymous
            
    Finally c)(V_0*(R_1+R_2))/R_1
    
                    Answered by
            Anonymous
            
    (Part c for RL circuit)
    
                    Answered by
            FLu
            
    Great many thanks Anonymous!
    
                    Answered by
            P
            
    Please tell  RL Circuit  I2 I3 for b and c
ASAP
    
ASAP
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