Post on 30-May-2017
transcript
• Series and parallel resistors• Kirchhoff’s Rules for network problems• Electrical meters and household circuits• R-C circuits
DC CIRCUITS: Chapter 26
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
Series connection
Parallel connection
Resistors connected in a circuit in series or parallel can be simplified using the
following:
C 1998 McDermott, et al., Prentice Hall
Calculate the total resistance of each box if each bulb has a resistance of 1000 Ohms ().
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 , 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…
Seewww.physics.sjsu.edu/Becker/physics51
These complex circuits cannot be reduced to series – parallel combinations.
So use Kirchhoff’s Rules:1. Ij = 0 junction rule
(valid at any junction); conservation of charge
2. (Vj ) = 0 loop rule(valid for any closed loop); conservation of energy
At node A, Iin = IoutI1 + I3 = I2
Vrises = VdropsLoop #1:I2R2 +4+ I1R1 = 1 + 2
Loop #2:3 + 2 = I2R2 + I3R3
• Label: 3 I’s; +/- on R’s; loops.
• Write equations.
Figure 26.66
5.00 A = I44.0
0B
I3 I2
Label the 3 branch currents I2, I3, and I4.Since VAB across all 3 branches is the same
and is known: V4 = I4R4 = 5A (4) = 20 Volts, the currents and can be readily solved.
A
V4 = I4R4 = 5A (4) = 20 Volts I3 = V3 / R3 = 4V / 3= 1.33 A
At junction B, I in = I outI4 = I2 + I3 ; I2 = I4 – I3 = 5A - 1.3A
I2 = 3.7ALoop #1: V rises = V drops = I2R2 + I4R4 = 3.7A (2) + 5A(4)
= 27.4 V
V4 = I4R4 = 5A (4) = 20 Volts I3 = V3 / R3 = 4V / 3= 1.33 AAt junction B, I in = I out
I4 = I2 + I3 ; I2 = I4 - I3 = 5A - 1.3A = 3.7ALoop #1: V rises = V drops
= I2R2 +I4R4 = 3.7A (2) + 5A(4) =27.4 V
5.0 A4.0
0B
I3 I2
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 (V) have a very large series resistor in them to reduce the amount of current drawn from the circuit being measured.
•DISCHARGING: CHARGING:
•An RC circuit that can be
used to charge and discharge
a capacitor (through a resistor).
•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
•A battery, a capacitor, and a resistor are connected in series. Which of the following affect(s) the maximum charge stored on the capacitor?
•Q26.26
•A. the emf of the battery
•B. the capacitance C of the capacitor
•C. the resistance R of the resistor
•D. both and C
•E. all three of , C, and R
House wiring circuits
Hand drill circuit with ground wire for safety
see www.physics.sjsu.edu/Becker/physics51
Review
C 2012 J. F. Becker