Solution to #1 of HUANG fall 07 midterm II
We can reduce the circuit to a single loop by successively
combining parallel and series combinations.
We combine
R
1
and
R
2
, which are in series:
R
7
=
R
1
+
R
2
= 2.8 k
Ω
+ 2.8 k
Ω
= 5.6 k
Ω
.
We combine
R
3
and
R
7
, which are in parallel:
1/
R
8
= (1/
R
3
) + (1/
R
7
) = [1/(2.8 k
Ω
)] + [1/(5.6 k
Ω
)],
which gives
R
8
= 1.87 k
Ω
.
We
combine
R
4
and
R
8
, which are in series:
R
9
=
R
4
+
R
8
= 2.8 k
Ω
+ 1.87 k
Ω
= 4.67 k
Ω
.
We combine
R
5
and
R
9
, which are in parallel:
1/
R
10
= (1/
R
5
) + (1/
R
9
) = [1/(2.8 k
Ω
)] + [1/(4.67 k
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This note was uploaded on 02/04/2010 for the course PHYSICS 7B taught by Professor Packard during the Spring '08 term at Berkeley.
 Spring '08
 Packard

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