Comments for The Pannist http://thepannist.com/blog a physicist? maybe... or may be not. Wed, 23 Sep 2009 17:13:19 +0100 http://wordpress.org/?v=2.8.4 hourly 1 Comment on Selected Problems on DC-circuit by johannes http://thepannist.com/blog/?p=152&cpage=1#comment-21 johannes Wed, 23 Sep 2009 17:13:19 +0000 http://thepannist.com/blog/?p=152#comment-21 The result for A in problem 3013 is wrong! It must be [tex]A = \frac{R_1+R_3}{(R_1 R_2+R_3 R_2+R_3 R_1) C}[/tex] with best regards The result for A in problem 3013 is wrong!

It must be

A = \frac{R_1+R_3}{(R_1 R_2+R_3 R_2+R_3 R_1) C}

with best regards

]]> Comment on Selected AC-circuit problems from Feynman’s Lectures on Physics Exercise Vol 2 by Pannini http://thepannist.com/blog/?p=137&cpage=1#comment-20 Pannini Wed, 20 Aug 2008 08:14:16 +0000 http://thepannist.com/blog/?p=137#comment-20 solution to 22-3 (b) We can take a result from 22-6(a) that the impedance of the circuit is given by [tex]{\frac {i\omega\,L + {i}^{2}{\omega}^{2}LRC + R}{1 + 2\,Ri\omega\,C + {i}^{2}{\omega}^{2}CL}}[/tex] at [tex]{\omega}^{2}=\frac{1}{LC}[/tex], this reduces to [tex]\frac{L}{2CR}[/tex] In other words, if the source has an amplitude V, the current flowing in the circuit would be [tex]\frac{2CRV}{L}[/tex] and the power delivered by the source would be [tex]\frac{2CRV^2}{L}[/tex] Back to the simplified circuit, the voltage across the left hand side is V, the voltage across R on the left hand side is therefore given by [tex]\frac{ \left( {R}^{-1} + i\omega\,C \right)^{-1}V}{ \left( i\omega\,L + \left( {R}^{-1}+ i\omega\,C \right) ^{-1} \right) }[/tex] After simplification and substitution of [tex]{\omega}^{2}=\frac{1}{LC}[/tex], we find that the voltage across this R is simply [tex]\frac{VR}{i\omega L}[/tex] Similarly, we can also do this to find the voltage across R on the right hand side of the circuit. Surprisingly, the voltage across that R is [tex]\frac{-VR}{i\omega L}[/tex] Power dissipation on these two resistors are then [tex]\frac{2RV^2}{i\omega^2 L^2}[/tex] or [tex]\frac{RV^2}{i\omega^2 L^2}[/tex] each Subsitute [tex]{\omega}^{2}=\frac{1}{LC}[/tex], we find that this equals to the power delivered by the AC source as found earlier. [tex]P=\frac{2CRV^2}{L}[/tex] solution to 22-3 (b)

We can take a result from 22-6(a) that the impedance of the circuit is given by
{\frac {i\omega\,L + {i}^{2}{\omega}^{2}LRC + R}{1 + 2\,Ri\omega\,C + {i}^{2}{\omega}^{2}CL}}
at {\omega}^{2}=\frac{1}{LC}, this reduces to
\frac{L}{2CR}
In other words, if the source has an amplitude V, the current flowing in the circuit would be \frac{2CRV}{L} and the power delivered by the source would be
\frac{2CRV^2}{L}

Back to the simplified circuit, the voltage across the left hand side is V, the voltage across R on the left hand side is therefore given by
\frac{ \left( {R}^{-1} + i\omega\,C \right)^{-1}V}{ \left( i\omega\,L + \left( {R}^{-1}+ i\omega\,C \right) ^{-1} \right) }
After simplification and substitution of {\omega}^{2}=\frac{1}{LC}, we find that the voltage across this R is simply
\frac{VR}{i\omega L}
Similarly, we can also do this to find the voltage across R on the right hand side of the circuit. Surprisingly, the voltage across that R is
\frac{-VR}{i\omega L}

Power dissipation on these two resistors are then
\frac{2RV^2}{i\omega^2 L^2} or \frac{RV^2}{i\omega^2 L^2} each

Subsitute {\omega}^{2}=\frac{1}{LC},
we find that this equals to the power delivered by the AC source as found earlier.
P=\frac{2CRV^2}{L}

]]> Comment on Selected AC-circuit problems from Feynman’s Lectures on Physics Exercise Vol 2 by Pannini http://thepannist.com/blog/?p=137&cpage=1#comment-14 Pannini Wed, 20 Aug 2008 06:12:07 +0000 http://thepannist.com/blog/?p=137#comment-14 solution to 22-3 (a) We can redraw the circuit. The simplified version would be a source connected to two main parts, which are connected in parallel. <code> |--------------------------- |               |          | |               L          C |               |          | |             -----      ----- ~             |   |      |   | |             R   C      L   R |             |   |      |   | |             -----      ----- |               |          | |--------------------------- </code> The complex impedance of the circuit is therefore given by [tex]\left( \left( i\omega\,L + \left( {R}^{-1}+ i\omega\,C \right) ^{-1} \right) ^{-1} + \left( {\frac {1}{i\omega\,C}} + \left( {R}^{-1} +{\frac {1}{i\omega\,L}} \right) ^{-1} \right) ^{-1} \right) ^{-1}[/tex] That's damn scary, isn't it? Ha Ha After a little simplification, that reduces to [tex]{\frac {i\omega\,L+ {i}^{2}{\omega}^{2}LRC +R}{1 +2\,Ri\omega\,C +{i}^{2}{\omega}^{2}CL}}[/tex] When we substitute $$R^2=\frac{L}{2C}$$ in, the imaginary part vanishes, the real part is left only by $$!R$$ Umm, very impressive.... solution to 22-3 (a)
We can redraw the circuit. The simplified version would be a source connected to two main parts, which are connected in parallel.


|---------------------------
|               |          |
|               L          C
|               |          |
|             -----      -----
~             |   |      |   |
|             R   C      L   R
|             |   |      |   |
|             -----      -----
|               |          |
|---------------------------

The complex impedance of the circuit is therefore given by
\left( \left( i\omega\,L + \left( {R}^{-1}+ i\omega\,C \right) ^{-1} \right) ^{-1} + \left( {\frac {1}{i\omega\,C}} + \left( {R}^{-1} +{\frac {1}{i\omega\,L}} \right) ^{-1} \right) ^{-1} \right) ^{-1}
That’s damn scary, isn’t it? Ha Ha
After a little simplification, that reduces to
{\frac {i\omega\,L+ {i}^{2}{\omega}^{2}LRC +R}{1 +2\,Ri\omega\,C +{i}^{2}{\omega}^{2}CL}}
When we substitute R^2=\frac{L}{2C} in, the imaginary part vanishes, the real part is left only by

R

Umm, very impressive….

]]> Comment on Selected AC-circuit problems from Feynman’s Lectures on Physics Exercise Vol 2 by Pannini http://thepannist.com/blog/?p=137&cpage=1#comment-9 Pannini Wed, 20 Aug 2008 04:52:19 +0000 http://thepannist.com/blog/?p=137#comment-9 Solution for 22-6 when there is no current flowing through , voltage on two sides are equal at all time. That can be obtained when the ratio of the voltage across $$R_{a}$$ and L to the whole left hand side and the voltage across and to the whole right hand side are equal. Hence, it can be written as [tex]\frac {R_{a} + i\omega\,L}{R_{a}+ i\omega\,L +R}=\frac {R}{R + \left( {R_{{b}}}^{-1} +i\omega\,C \right) ^{-1}}[/tex] which can be solved for L. [tex]L={\frac {-R_{{a}}R_{{b}}+ {R}^{2} +{R}^{2}i\omega\,CR_{{b}}}{i\omega\,R_{{b}}}}[/tex] Since L must be real, the complex part has to vanish, and we are left with the real part. $$!L=R^2C$$ Solution for 22-6

when there is no current flowing through , voltage on two sides are equal at all time. That can be obtained when the ratio of the voltage across R_{a} and L to the whole left hand side and the voltage across and to the whole right hand side are equal.
Hence, it can be written as
\frac {R_{a} + i\omega\,L}{R_{a}+ i\omega\,L +R}=\frac {R}{R + \left( {R_{{b}}}^{-1} +i\omega\,C \right) ^{-1}}
which can be solved for L.
L={\frac {-R_{{a}}R_{{b}}+ {R}^{2} +{R}^{2}i\omega\,CR_{{b}}}{i\omega\,R_{{b}}}}
Since L must be real, the complex part has to vanish, and we are left with the real part.

L=R^2C

]]> Comment on B7 IB Adv P1 2004 by Pannini http://thepannist.com/blog/?p=81&cpage=1#comment-6 Pannini Mon, 18 Aug 2008 04:19:20 +0000 http://thepannist.com/blog/?p=81#comment-6 Solution Due to the change in magnetic flux while the square-shaped loop is entering the field region, there would be an induced emf and hence, a current flowing in the wire. $$\frac{d\Phi }{dt}=Bab\omega=IR$$ This current exist while the loop is entering and emerging from the field region, but not when the loop is completely in the field region. As a result, the total power dissipated due to the resistance of the wire can be expressed as $$P\Delta t=IV\Delta t=2\frac{(Bab\omega)^2}{R}\frac{a}{\omega b}$$ In the mean time, the induced current interacts with the field, giving rise in the opposing force. $$F=ILB=\frac{Bab\omega}{R}aB$$ Work done by the field is $$W=F\Delta S=\frac{Bab\omega}{R}aB\times 2a$$ After a little rearrangement, it can be seen that the work done by the field is the same as power dissipated in the wire. Solution

Due to the change in magnetic flux while the square-shaped loop is entering the field region, there would be an induced emf and hence, a current flowing in the wire.
\frac{d\Phi }{dt}=Bab\omega=IR
This current exist while the loop is entering and emerging from the field region, but not when the loop is completely in the field region.
As a result, the total power dissipated due to the resistance of the wire can be expressed as
P\Delta t=IV\Delta t=2\frac{(Bab\omega)^2}{R}\frac{a}{\omega b}
In the mean time, the induced current interacts with the field, giving rise in the opposing force.
F=ILB=\frac{Bab\omega}{R}aB
Work done by the field is
W=F\Delta S=\frac{Bab\omega}{R}aB\times 2a
After a little rearrangement, it can be seen that the work done by the field is the same as power dissipated in the wire.

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