Transcript DC Circuits - San Jose State University

DC CIRCUITS: Chapter 26
• Series and parallel resistors
• Kirchhoff’s Rules for network problems
• Electrical meters and household circuits
• R-C circuits
C 2012 J. Becker
Learning Goals - we will learn:
• How to simplify resistors connected
in a circuit in series and in parallel.
• How to simplify and analyze more
complicated networks using
Kirchhoff’s Rules.
• R-C circuits
Resistors connected in a circuit in series or
parallel can be simplified using the following:
Series connection
Parallel connection
Calculate the total resistance of each box if
each bulb has a resistance of 1000 Ohms (W).
C 1998 McDermott, et al., Prentice Hall
Method of simplifying the circuit in (a) below
to get the equivalent resistance.
We can then calculate the power P = I2 R
dissipated in each resistor.
Exercise 26.13: Four equivalent light bulbs
R1 = R2 = R3 = R4 = 4.50 W, emf = 9.00 Volts.
Find current and power in each light bulb.
Which bulb is brightest?
Later, if bulb #4 is removed which bulbs
get brighter? Dimmer?
On the course website, see old Test #1
from previous semesters
for typical DC circuit problems
AND other typical problems on tests…
See
www.physics.sjsu.edu/Becker/physics51
These complex circuits cannot
be reduced to series –
parallel combinations.
So use Kirchhoff’s Rules:
1. S Ij = 0 junction rule
(valid at any junction);
conservation of charge
2. S (DVj ) = 0 loop rule
(valid for any closed loop);
conservation of energy
At node A, S Iin = S Iout
I1 + I 3 = I 2
S Vrises = S Vdrops
Loop #1:
• Label:
3 I’s;
+/- on R’s;
loops.
• Write equations.
I2R2 +e4+ I1R1 = e1 + e2
Loop #2:
e 3 + e 2 = I 2 R2 + I3 R3
Figure 26.66
A
I3 I2
5.00 A = I4
4.00
B
Label the 3 branch currents I2, I3, and I4.
Since VAB across all 3 branches is the same
and is known: V4 = I4R4 = 5A (4W) = 20 Volts,
the currents and
e can be readily solved.
V4 = I4R4 = 5A (4W) = 20 Volts
I3 = V3 / R3 = 4V / 3W = 1.33 A
At junction B,
S I in = S I out
I4 = I2 + I3 ; I2 = I4 – I3 = 5A - 1.3A
I2 = 3.7A
Loop #1:
S V rises = S V drops
e = I2R2 + I4R4 = 3.7A (2W) + 5A(4W)
e = 27.4 V
5.0 A
I3
4.00
I2
B
V4 = I4R4 = 5A (4W) = 20 Volts
I3 = V3 / R3 = 4V / 3W = 1.33 A
At junction B,
S I in = S I out
I4 = I2 + I3 ; I2 = I4 - I3 = 5A - 1.3A = 3.7A
Loop #1:
S V rises = S V drops
e = I2R2 +I4R4 = 3.7A (2W) + 5A(4W) =27.4 V
Figure 26.11
Typo on top of page 890 (12th Edition),
2nd line from top of page:
“right side” should read “bottom”
(See Figure 26.11 which is rotated 90o
from figure in the 11th Edition.)
ELECTRICAL MEASURING INSTRUMENTS –
METERS
A d’Arsonval galvanometer meter movement
AMMETERS have a very small shunt resistor in them
to reduce the effect of introducing the meter
resistance into the circuit being measured.
VOLTMETERS (DV) have a very large series resistor
in them to reduce the amount of current drawn from
the circuit being measured.
•DISCHARGING:
•An RC circuit
that can be used
to charge and
discharge a
capacitor
(through a
resistor).
CHARGING:
•CHARGING A CAPACITOR:
current vs time
•CHARGING A CAPACITOR:
charge vs time
•DISCHARGING A CAPACITOR:
current vs time
•DISCHARGING A CAPACITOR:
charge vs time
o
o
•Q26.26
•A battery, a capacitor, and a resistor are connected in series.
Which of the following affect(s) the maximum charge stored
on the capacitor?
•A. the emf e of the battery
•B. the capacitance C of the capacitor
•C. the resistance R of the resistor
•D. both e and C
•E. all three of e, C, and R
House wiring circuits
Hand drill circuit with ground wire for
safety
Review
see
www.physics.sjsu.edu/Becker/physics51
C 2012 J. F. Becker