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MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics
8.02
Spring 2010
Problem Set 6 Solutions
Problem 1: Four Resistors
Four resistors are connected to a battery as shown in the
figure. The current in the battery is
I
, the battery emf is
ε
, and
the resistor values are
R
1
=
R
,
R
2
= 2
R
,
R
3
= 4
R
,
R
4
= 3
R
.
(a) Rank the resistors according to the potential difference
across them, from largest to smallest. Note any cases of equal
potential differences.
Resistors 2 and 3 can be combined (in series) to give
23
2
3
246
R
RR
R
R
R
=+=
+
=
.
23
R
is in parallel with
4
R
and
the equivalent resistance
234
R
is
23
4
234
23
4
(6 )(3 )
2
63
R
R
==
=
++
Since
234
R
is in series with
1
R
, the equivalent resistance of the whole circuit is
1234
1
234
23
R
RR R R R
=+
=
+
=
. In series, potential difference is shared in proportion to
the resistance, so
1
R
gets 1/3 of the battery voltage (
1
/3
V
ε
Δ
=
) and
234
R
gets 2/3 of the
battery voltage (
234
2/
3
V
Δ=
). This is the potential difference across
4
R
(
4
3
V
),
but
2
R
and
3
R
must share this voltage: 1/3 goes to
2
R
(
2
(
1
/
3
)
(
3
) 2/
9
V
Δ
) and
2/3
to
3
R
(
3
(2/3)(2 /3)
4 /9
V
=
) . The ranking by potential difference is
4312
VVVV
Δ>
Δ
.
(b) Determine the potential difference across each resistor in terms of
ε
.
As shown from the reasoning above, the potential differences are
1
234
242
,,
,
39
9
3
VV
V
V
εεε
Δ= Δ=
(c) Rank the resistors according to the current in them, from largest to smallest. Note any
cases of equal currents.
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All the current goes through
1
R
, so it gets the most (
1
I
I
=
). The current then splits at the
parallel combination.
4
R
gets more than half, because the resistance in that branch is less
than in the other branch.
2
R
and
3
R
have equal currents because they are in series. The
ranking by current is
1423
I
III
>>=
.
(d) Determine the current in each resistor in terms of
I
.
1
R
has a current of
I
. Because the resistance of
2
R
and
3
R
in series
(
23
2
3
246
R
RR
R
R
R
=+=
+
=
) is twice that of
4
3
R
R
=
, twice as much current goes
through
4
R
as through
2
R
and
3
R
. The current through the resistors are
12
3
4
2
,,
33
I
I
II
I
==
=
=
(e) If
R
3
is increased, what happens to the current in each of the resistors?
Since
23
4
2
3
4
1234
1
234
1
1
23
4
2
3
4
()
R
R
R
R
R
R
R
R
R
+
=+
+
++
increasing
3
R
increases the equivalent resistance of the entire circuit. The current in the
circuit, which is the current through
1
R
, decreases. This decreases the potential difference
across
1
R
, increasing the potential difference across the parallel combination. With a
larger potential difference the current through
4
R
is increased. With more current going
through
4
R
, and less in the circuit to start with, the current through
2
R
and
3
R
must
decrease. Thus,
41
2
3
increases and
,
, and
decrease
I
I
(f) In the limit that
R
3
→
∞
, what are the new values of the current in each resistor in
terms of
I
, the original current in the battery?
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 Spring '10
 DOURMASHKIN
 Physics, Current, Mass

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