Transcript Chapter 28

Chapter 28
Direct Current Circuits (Cont.)
Dr. Jie Zou
PHY 1361
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Outline
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Kirchhoff’s rules (28.3)
RC circuits (28.4)
Electrical meters (28.5)
Dr. Jie Zou
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Kirchhoff’s rules
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Kirchhoff’s rules are used to simplify the
procedure for analyzing more complex circuits:
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2.
Junction rule. The sum of the currents entering
any junction in a circuit must equal the sum of the
currents leaving that junction:  Iin =  Iout
Loop rule. The sum of the potential differences
across all elements around any closed circuit loop
must be zero:  V  0
closed
loop
Dr. Jie Zou
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Sign conventions
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Note the following sign conventions when
using the 2nd Kirchhoff’s rule:
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Each circuit element is assumed 
to be traversed from left to
right.
Dr. Jie Zou
If a resistor is traversed in the direction of the
current, the potential difference V across the
resistor is –IR (Fig. a).
If a resistor is traversed in the direction
opposite the current, the potential difference
V across the resistor is +IR (Fig. b).
If a source of emf is traversed in the direction
of the emf (from – to +), the potential
difference V is + (Fig. c).
If a source of emf is traversed in the direction
opposite the emf (from + to -), the potential
difference V is - (Fig. d).
Note: We have assumed that battery has no
internal resistance.
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Example 28.7 A single-loop
circuit
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A single-loop circuit contains two
resistors and two batteries, as
shown in the figure below.
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(A) Find the current in the circuit.
(Answer: -0.33 A)
(B) What power is delivered to each
resistor? What power is delivered by
the 12-V battery? (Answer: 0.87 W,
1.1 W, 4.0 W)
Note: The “-” sign for I indicates that
the direction of the current is
opposite the assumed direction.
See P. 879 on the textbook for the
Problem-solving hints.
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Example 28.8 A multi-loop
circuit
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Find the currents I1, I2, and I3 in the
circuit shown in the figure. (Answer:
I1= 2.0 A, I2= -3.0 A, I3= -1.0 A)
Problem-solving skills:
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Choose a direction for the current in
each branch.
Apply the junction rule.
Choose a direction (clockwise or
counterclockwise) to transverse each
loop.
Apply the loop rule.
In order to solve a particular circuit
problem, the number of independent
equations you need to obtain from the
two rules equals the number of
unknown currents.
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Charging a capacitor
t<0
t>0
q(t) = C(1 – e-t/RC)
= Q (1 – e-t/RC)
I(t) = dq/dt = (/R)e-t/RC
 = RC: time constant of
the circuit.
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Examples
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Quick Quiz: Consider the circuit in
the figure and assume that the
battery has no internal resistance.
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Just after the switch is closed, the
current in the battery is (a) zero (b)
/2R (c) 2/R (d) /R (e) impossible
to determine.
After a very long time, the current in
the battery is (f) zero, (g) /2R (h)
2/R (i) /R (j) impossible to
determine.
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Discharging a capacitor
q(t) = Qe-t/RC
I(t) = dq/dt = -(Q/RC)e-t/RC
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Example 28.12 Discharging a
capacitor in an RC circuit
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Consider a capacitor of capacitance C
that is being discharged through a
resistor of resistance R.
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After how many time constants is the
charge on the capacitor one-fourth its
initial value? (answer: 1.39 )
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Electrical meters
Voltmeter
Ammeter
Operation principle of
a galvanometer
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