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Unformatted text preview: than R3 + R4
since the branch containing
R5 has less resistance. R1 V R2 R4 R3
R5 ConcepTest 4.7 Junction Rule
ConcepTest
Junction
a) 2 A
What is the current in branch P? b) 3 A
c) 5 A
d) 6 A
e) 10 A 5A P 8A
2A ConcepTest 4.7 Junction Rule
Junction
a) 2 A
b) 3 A What is the current in branch P? c) 5 A
d) 6 A
e) 10 A The current entering the junction
in red is 8 A, so the current
red
leaving must also be 8 A. One
exiting branch has 2 A, so the
exiting
other branch (at P) must have 6
A. S 5A P 8A
junction 2A 6A ConcepTest 4.9 Wheatstone Bridge
ConcepTest
An ammeter A is connected
An a) I
a) between points a and b in the b) I/2
b) I/2 circuit below, in which the four c) I/3
c) I/3 resistors are identical. The current
identical The d) I/4
d) I/4 through the ammeter is:
through e) zero a b
V I ConcepTest 4.9 Wheatstone Bridge
ConcepTest
Wheatstone
An ammeter A is connected
An a) I
a) between points a and b in the b) I/2
b) I/2 circuit below, in which the four c) I/3
c) I/3 resistors are identical. The current
identical The d) I/4
d) I/4 through the ammeter is:
through e) zero Since all resistors are identical,
resistors a the voltage drops are the same
voltage
across the upper branch and the
lower branch. Thus, the
potentials at points a and b are
potentials b also the same. Therefore, no
same
current flows. V I ConcepTest 4.10
ConcepTest More Kirchhoff’s Rules
Kirchhoff’s
a) 2 – I1 – 2I2 = 0 Which of the equations is valid
Which b) 2 – 2I1 – 2I2 – 4I3 = 0
b)
2I
4I for the circuit below? c ) 2 – I 1 – 4 – 2 I2 = 0
c)
d ) I 3 – 4 – 2 I2 + 6 = 0
d)
e) 2 – I1 – 3I3 – 6 = 0
e)
1Ω
I2 2Ω
6V 2V
2V
4V I1
1Ω I3
3Ω ConcepTest 4.10 More Kirchhoff’s Rules
Kirchhoff’s
a) 2 – I1 – 2I2 = 0 Which of the equations is valid b) 2 – 2I1 – 2I2 – 4I3 = 0
b)
2I
4I for the circuit below? c ) 2 – I 1 – 4 – 2 I2 = 0
c)
d ) I 3 – 4 – 2 I2 + 6 = 0
d)
e) 2 – I1 – 3I3 – 6 = 0
e) Eqn. 3 is valid for the left loop:
Eqn.
The left battery gives +2V, then
there is a drop through a 1Ω
resistor with current I1 flowing.
Then we go through the middle
battery (but from + to – !), which
gives –4V. Finally, there is a
drop through a 2Ω resistor with
current I2. 1Ω
I2 2Ω
6V 2V
2V
4V I1
1Ω I3
3Ω...
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 Spring '11
 Young
 Physics

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