Chapter 25
Energy and
Current in DC Circuits:
(Additional Questions)
1)
In figure, if the potential at point
P
is 100 V, what is the potential at point Q?
Solution:
The current in the circuit is
So from
we get
2) (a)
In figure, what is the equivalent resistance of the network shown?
(b)
What is the
current in each resistor? Put
1
100
R
=
Ω
,
2
3
50
R
R
=
=
Ω
,
4
75
R
=
Ω
, and
E
=6 V; assume the
battery is ideal.
Solution:
(a)
R
2
,
R
3
and
R
4
are in parallel. By finding a common denominator and simplifying, the
equation 1
/R
= 1
/R
2
+ 1
/R
3
+ 1
/R
4
gives an equivalent resistance of
2
3
4
2
3
2
4
3
4
(50
)(50
)(75
)
18.75
.
(50
)(50
)
(50
)(75
)
(50
)(75
)
R R R
R
R R
R R
R R
Ω
Ω
Ω
=
=
=
Ω
+
+
Ω
Ω +
Ω
Ω +
Ω
Ω
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View Full DocumentThus, considering the series contribution of resistor
R
1
, the equivalent resistance for the
network is
1
100
18.75
118.75
.
eq
R
R
R
=
+
=
Ω +
Ω =
Ω
(b)
1
eq
i
R
=
E
6 V
0.0505 A
118.75
=
=
Ω
;
(
29
(
29
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 Spring '09
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 Resistor, Jaguar Racing, Electrical resistance

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