page 34
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C1 |
R1 |
Vs |
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C2 |
I1 |
L1 |
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I2 |
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For the two loops, |
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∑ VI1 |
= – VS + C1∫ I1dt + C2∫ ( I1 – I2) dt = 0 |
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VS = ( C1 + C2) ∫ I1dt–C2∫ I2dt |
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VS + C2∫ I2dt |
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∫ I1dt = ------------------------------- |
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C1 + C2 |
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∑ VI2 |
= C2∫ ( I2 |
– I1) dt + R1I2 |
+ L |
d |
I2 = 0 |
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dt |
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sub. (1) into (2), |
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VS + C2∫ I2dt |
∫ I2dt + R1I2 |
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d |
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–C2 |
------------------------------- + C2 |
+ L |
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I2 |
= 0 |
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C1 + C2 |
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dt |
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(1)
(2)
etc.......
4. PASSIVE DEVICES
• Passive devices will have the same operating characteristics at the same operating points.
page 35
4.1 TRANSFORMERS
•A transformer can be viewed as a converter that can increase voltage and lower current, or vice versa. It only works when using AC.
•The transformer is effectively a magnetic circuit. The transformer has two or more coils of wire wrapped about a common core.
II
+
VI
-
• The ideal relationship is,
II |
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N1:N2 |
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Io |
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VI |
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Vo |
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II |
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N1:N2 |
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Io |
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VI |
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Vo |
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+ |
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Io
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Vo
-
VI |
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Vo |
and |
IINI = IoNo |
----- |
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NI |
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No |
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where,
N1 = the number of coils on the primary side N2 = the number of coils on the secondary side
• If a transformer has an iron core it will be shown with lines in the centre,
page 36
•To deal with a transformer in a circuit analysis we need to pay attention to the polarity of the coils, and we may consider the inductance of each coil at times.
•Consider the example below, from [Nilsson, pg. 450]. We want to find the power delivered to the 1ohm resistor. We will use the mesh current method,
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4Ω |
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I2 |
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7.2Ω |
4:1 |
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120V (RMS) |
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1Ω |
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I3 |
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I1 |
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For I1, |
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120V = 7.2( I1 – I2) |
+ Vt |
(1) |
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For I2,
4I2 |
Vt |
– Vt |
+ 7.2( I2 – I1) |
= |
0 |
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(2) |
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+ ---- |
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For I3, |
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4 |
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1I |
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Vt |
= 0 |
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V |
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= 4I |
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(3) |
3 |
– ---- |
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3 |
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4 |
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(3) into (1), |
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120 |
= 7.2I1 – 7.2I2 + 4I3 |
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(3) into (2), |
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0 = |
– 7.2I1 + 11.2I2 – 3I3 |
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For the transformer, |
4( I1 – I2) = 1( I3 – I2) |
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I1 – 3I2 – I3 |
= 0 |
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(4) |
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