Electronics - Inductors - Discussion
Discussion Forum : Inductors - General Questions (Q.No. 2)
2.
In the given circuit, what will the voltage be across R3 25 
s after the switch is moved to position 2?
Discussion:
18 comments Page 1 of 2.
Isuru_R said:
1 decade ago
When the switch is at "1" position
I = 14/(120 + 820)
I = 0.01489
Then we can find the current 25us after the switch is moved to position "2" using the equation I2 = I * e^(-tR/L)
Then... from(R3* I2 ) we get the voltage across the resistor "R3".
Substitute I = 0.01489 A ; t = 25x10^(-6) s ; R = (820 + 560) ;
and L = 0.1 H
Then you will get (I2*R3) = 5.9 V , which gaves us the required solution....!
I = 14/(120 + 820)
I = 0.01489
Then we can find the current 25us after the switch is moved to position "2" using the equation I2 = I * e^(-tR/L)
Then... from(R3* I2 ) we get the voltage across the resistor "R3".
Substitute I = 0.01489 A ; t = 25x10^(-6) s ; R = (820 + 560) ;
and L = 0.1 H
Then you will get (I2*R3) = 5.9 V , which gaves us the required solution....!
(1)
Pritam said:
5 years ago
I(0-)=14/940=0.0148A....Inductor gets shorted.
At position 2.
inductor starts discharging
I(t)=I(0-)*e^(-t/T)---------> (a)
T=L/R=.1/1380=7.2*10^-5.
t/T=25*10^-6/7.2*10^-5=0.3472.
put this value in (a).
I(25us)=0.01052.
The voltage of R3=560*0.01052=5.9V.
At position 2.
inductor starts discharging
I(t)=I(0-)*e^(-t/T)---------> (a)
T=L/R=.1/1380=7.2*10^-5.
t/T=25*10^-6/7.2*10^-5=0.3472.
put this value in (a).
I(25us)=0.01052.
The voltage of R3=560*0.01052=5.9V.
(1)
Oliver said:
6 years ago
Yes, but to be clear, inductors are shorted (VL = 0) at DC only when STEADY STATE is reached (around 5 time constants). Though the question did not mention it, it was assumed to be that way which is why its more confusing.
Syed Shams Ul Arifeen Shah said:
9 years ago
@Mayank,
Yes we ignore the inductance because in DC source, frequency is zero so that inductor behave like short wire (XL = 2 * pi * F * L) here F = 0, XL = 0.
Yes we ignore the inductance because in DC source, frequency is zero so that inductor behave like short wire (XL = 2 * pi * F * L) here F = 0, XL = 0.
Ashfaque said:
7 years ago
@All.
An Inductor is a short circuit to dc. Then why we would be bothered about 25 microseconds. I think you are correct @Naseem.
An Inductor is a short circuit to dc. Then why we would be bothered about 25 microseconds. I think you are correct @Naseem.
Priyanka said:
6 years ago
@Clint.
I2 = I*(e^(t(R/L)).
Then I2 = 10.54*10^-3.
Therefore voltage across R3 = I2*R3 =5.905 v.
I2 = I*(e^(t(R/L)).
Then I2 = 10.54*10^-3.
Therefore voltage across R3 = I2*R3 =5.905 v.
Naseem ul aziz said:
1 decade ago
Apply voltage divider that is
when switch will be at position 2
v3=((r3)/(r3+r2))*vt=5.6 volt
when switch will be at position 2
v3=((r3)/(r3+r2))*vt=5.6 volt
Mayank said:
9 years ago
Did we ignore inductance when the switch is at position one due to its very less magnitude?
Clint said:
1 decade ago
@Isuru, @Sambhaji, @Preetha, could you please solve I2 ? I didn't get the correct answer.
Chandhu said:
9 years ago
Open switch are there r1 then why voltage flow r3?
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