HW 5
1
(a)
The 1 uf and 2 uf in series on the left have
a cap of (1)(2)/(1+2) = 2/3.
The 2/3 uf in
parallel with the 3 uf is (2/3)+3 = 3.6666 uf.
Going to the right side of the circuit, the 1 uf
and 1 uf in parallel have an equivalent
capacitance of 1+1 = 2 uf.
The 2 uf in series
with the 2 uf on the right has an equivalent
capacitance of (2)(2)/(2+2) = 1 uf.
The
remaining 3.6666 uf and 1 uf are in parallel so
their capacitance is the sum or
4.6666 uf.
(b)
The 2 uf and 1 uf in parallel have 3 uf
total.
The 4 uf and 2 uf in parallel have 6 uf
total.
The 6 uf in series with the 3 uf have (6)
(3)/(6+3) =
2 uf
which is the answer.
(a)
The 1 h and 3 h are in series so the total
for them is 4 h.
The 4 h is in parallel with the
2 h so its equivalent inductance is (4)(2)/
(4+2)=8/6 = 1.3333 h.
The 1.3333 h is in
series with the remaining 1 h inductance so the
total is their sum or
2.3333 h
.
(b)
The 2 h and 2 h are in parallel so their
equivalent is (2)(2)/(2+2)=1 h.
The 2 h and 6
h are in parallel so their equivalent is (2)(6)/
(2+6)=12/8 = 1.5 h.
The 1.5 h is in series with
the 1 h so that inductance is 2.5 h that is to the
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 Fall '08
 Staff
 Resistor, Volts

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