- #1
issacnewton
- 1,000
- 29
Hello
I am reading Young, Freedman's University physics and I have some questions about the
sign conventions followed in getting the equations for the current decay in the R-L circuit.
I have attached two snapshots which will help the discussions. Consider the series R-L circuit
as shown in the diagram. Say switch S1 is closed and switch S2 is open for a long time. So that means steady current has been established in the upper part of
the circuit. Now open S1 and close S2. Now authors give the following
equation for the first case, when the current is building.
[tex]\mathcal{E}-iR -L\frac{di}{dt}=0[/tex]
which makes sense using the second diagram I attached. But the authors say that , for the
second case , when the current is decaying through the bottom part of the circuit, the equation to be solved becomes,
[tex]-iR -L\frac{di}{dt}=0[/tex]
which doesn't make sense. Since the current is now decreasing, going around the loop
in the direction of current (assuming that the direction of the current is still the same),
and using the second diagram which I have uploaded, the Kirchhoff's rule says that
[tex] -iR +L\frac{di}{dt}=0[/tex]
But this doesn't give correct final result for the decay. So am I doing something wrong ?
thanks
I am reading Young, Freedman's University physics and I have some questions about the
sign conventions followed in getting the equations for the current decay in the R-L circuit.
I have attached two snapshots which will help the discussions. Consider the series R-L circuit
as shown in the diagram. Say switch S1 is closed and switch S2 is open for a long time. So that means steady current has been established in the upper part of
the circuit. Now open S1 and close S2. Now authors give the following
equation for the first case, when the current is building.
[tex]\mathcal{E}-iR -L\frac{di}{dt}=0[/tex]
which makes sense using the second diagram I attached. But the authors say that , for the
second case , when the current is decaying through the bottom part of the circuit, the equation to be solved becomes,
[tex]-iR -L\frac{di}{dt}=0[/tex]
which doesn't make sense. Since the current is now decreasing, going around the loop
in the direction of current (assuming that the direction of the current is still the same),
and using the second diagram which I have uploaded, the Kirchhoff's rule says that
[tex] -iR +L\frac{di}{dt}=0[/tex]
But this doesn't give correct final result for the decay. So am I doing something wrong ?
thanks