An inductor with a value of 12 H is placed in serieswith a 4.7 kOhm resistor. A

Question

An inductor with a value of 12 H is placed in serieswith a 4.7 kOhm resistor. A 12 V potential is applied to thepair. A) What is the time constant () of the circuit? B) At what time does the current through the resistor reachhalf its maximum value? C) What is the current through the resistor att=0.5?
For part A, I used L= L/R=2.5532x10-9
For part B, I think that I will need to manipulate theequation i=(/R)(1-e-t/) but I am not surewhich side needs to be manipulated in order to find half thecurrent value.
For part C, I used theequation i=(/R)(1-e-(o.5)/), butI calculated a value of -0.001656 A, which I am pretty sure is notcorrect (due to the negative value).
Any assistance would be much appreciated. Thank you. A) What is the time constant () of the circuit? B) At what time does the current through the resistor reachhalf its maximum value? C) What is the current through the resistor att=0.5?
For part A, I used L= L/R=2.5532x10-9
For part B, I think that I will need to manipulate theequation i=(/R)(1-e-t/) but I am not surewhich side needs to be manipulated in order to find half thecurrent value.
For part C, I used theequation i=(/R)(1-e-(o.5)/), butI calculated a value of -0.001656 A, which I am pretty sure is notcorrect (due to the negative value).
Any assistance would be much appreciated. Thank you.

Explanation / Answer

You are on the right track again. A) What is the time constant () of the circuit? i(t) = (Vb/R)( 1 - e -t/ ) where =L/R= 12x10-6/ 4.7x103 = = 2.55 x10-9s B) At what time does the current through the resistor reachhalf its maximum value? Since i(t) = (Vb/R)( 1 - e -t/) and then since current through the resistor reach half itsmaximum value we have 0.5 = ( 1 - e -t/ ) t= ln(2) t= 2.55 x10-9 ln(2)   t=  1.77x10-9 s C) What is the current through the resistor att=0.5? again we use i(t) = (Vb/R)( 1 - e -t/ )and since t= 0.5
i(0.5) = (12/4.7x103 )( 1 - e-0.5 ) i(0.5) = 1.0 x10-3A A) What is the time constant () of the circuit? i(t) = (Vb/R)( 1 - e -t/ ) where =L/R= 12x10-6/ 4.7x103 = = 2.55 x10-9s B) At what time does the current through the resistor reachhalf its maximum value? Since i(t) = (Vb/R)( 1 - e -t/) and then since current through the resistor reach half itsmaximum value we have 0.5 = ( 1 - e -t/ ) t= ln(2) t= 2.55 x10-9 ln(2)   t=  1.77x10-9 s C) What is the current through the resistor att=0.5? again we use i(t) = (Vb/R)( 1 - e -t/ )and since t= 0.5
i(0.5) = (12/4.7x103 )( 1 - e-0.5 ) i(0.5) = 1.0 x10-3A

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---