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16. Line 1 has slope
R
1
= 6.0 k
Ω
.
Line 2 has slope
R
2
= 4.0 k
Ω
.
Line 3 has slope
R
3
=
2.0 k
Ω
.
The parallel pair equivalence is
R
12
=
R
1
R
2
/(
R
1
+
R
2
) = 2.4 k
Ω
.
That in series with
R
3
gives an equivalence of
123
12
3
2.4 k
2.0 k
4.4 k
.
RR
R
=
+=
Ω
+
Ω
=
Ω
The
current
through the battery is therefore
123
/
iR
ε
=
=
(6 V)/(4.4 k
Ω
) and the voltage drop across
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics, Current

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