PHYS 222
Worksheet 11 – Energy and Power
in Circuits
Supplemental Instruction
Iowa State University
Leader: Alek Jerauld
Course: PHYS 222
Instructor: Dr. Paula Herrera-Siklódy
Date: 2/9/12
Useful Equations
Req  R1  R2  R3  ...  RN
Equivalent resistance for resistors in series
1
1 1 1
1
Equivalent resistance for resistors in parallel
    ... 
Req R1 R2 R3
RN
Voltage across 2 resistors in series with a battery
Vs  V1 V2
Voltage across 2 resistors in parallel with a battery
Vs  V1  V2
Current through 2 resistors in series with a battery
I s  I1  I 2
Current through 2 resistors in parallel with a battery
I s  I1  I 2
R  R0 1  T  T0 


Resistance with respect to temperature. α is the
Temperature coefficient of resistance. Use absolute
temperatures.
P  VI  I 2 R 
V2
R
Power into a resistor/element in circuit
Related Problems
1) Consider the circuit shown in the figure. The terminal voltage of the 24.0-V battery is
21.2 V. The current in the circuit is 4.00 A. (Book 25.32)
(a) What is the internal resistance r of the battery?
Vr   V2  24  24.8  2.8 V
(b) What is the resistance R of the circuit resistor?
VR  21.2V  R 
VR 21.2

 5.3 
I
4
2) The circuit shown in the figure contains two batteries, each with an emf and an internal
resistance, and two resistors. (Book 25.36)
(a) Find the magnitude and direction of the current in the circuit
 eq  16  8  8 V
Req  1.6  5  1.4  9  17 
I
 eq
Req

8
 0.47 A
17
(b) Find the terminal voltage Vab of the 16-V battery
VT  16  (0.47)(1.6)  15.25 V
(c) Find the potential difference Vac of point a with respect to point c
Va,c  Va Vc  8 1.4(0.47)  5(0.47)  11 V
3) An idealized voltmeter is connected across the terminals of a 15.0-V battery, and a 75.0Ω appliance is also connected across its terminals. If the voltmeter reads 11.3 V. How
much power is being dissipated by the appliance? What is the internal resistance of the
battery? (Book 25.52)
+
V
r
-
e =15 V
R=75
VR  11.3 V
VR 2 11.32

 1.70 W
R
75
V
I  R  0.151 A
R
V  VR 15 11.3
r  r 

 26.5 
I
I
0.151
PR 
4) An idealized ammeter is connected to a battery as shown in the figure (Book 25.34)
(a) Find the reading of the ammeter
 10
I   5 A
r 2
(b) Find the current through the 4.00-Ω resistor
I 0 A
(c) Find the terminal voltage of the battery
V 0V
5) When switch S in the figure is open, the voltmeter V of the battery reads 3.11 V. When
the switch is closed, the voltmeter reading drops to 2.96 V, and the ammeter A reads
1.65 A. Assume that the two meters are ideal, so they don't affect the circuit. (Book 25.37)
(a) Find the emf
Since the switch is open no current goes through the
lower branch. The voltmeter is ideal, so no current
runs through the meter, thus current through internal
resister is zero. This means there is no voltage loss
across the internal resistor:
V1  3.11V
   3.11 V
(b) Find the internal resistance of the battery
V2  2.96V
I  1.65 A
V2    rI  r 
 V2
I
 0.09 
(c) Find the circuit resistance R
V2 is also the voltage across R:
R
V2
 1.79 
I
6) A typical small flashlight contains two batteries, each having an emf of 1.50 V connected
in series with a bulb having a resistance of 17 Ω. (Book 25.54)
(a) If the internal resistance of the batteries is negligible, what power is delivered to the
bulb?
P
 eq 2
R
 0.53 W
(b) If the batteries last for a time of 5.1 h, what is the total energy delivered to the bulb?
E  Ptsec onds  (0.53)(5.1)(3600)  9700 J
(c) The resistance of real batteries increases as they run down. If the initial internal
resistance is negligible, what is the combined internal resistance of both batteries when the
power to the bulb has decreased to half its initial value? (Assume that the resistance of the
bulb is constant. Actually, it will change somewhat when the current through the filament
changes, because this changes the temperature of the filament and hence the resistivity of
the filament wire.)
(r is the total internal resistance. R is the original resistor )
P'R 
P0
2
1 ( eq )
2 R
2
 

( )2
  eq  R  eq
2R
rR
1
1


2
2
 r  R  2R
2
 I f 2R 
r 


2 1 R  7.04 