Transcript EMF and Terminal P.D.

Chapter 28B - EMF and Terminal P.D.
A PowerPoint Presentation by
Paul E. Tippens, Professor of Physics
Southern Polytechnic State University
©
2007
Objectives: After completing this
module, you should be able to:
• Solve problems involving emf, terminal
potential difference, internal resistance,
and load resistance.
• Solve problems involving power gains and
losses in a simple circuit containing internal
and load resistances.
• Work problems involving the use of
ammeters and voltmeters in dc circuits.
EMF and Terminal
Potential Difference
The emf E is the open-circuit potential difference.
The terminal voltage VT for closed circuit is
reduced due to internal resistance r inside source.
Open
Circuit E = 1.5 V
Closed
Circuit VT = 1.45 V
r
Applying Ohm’s law
to battery r, gives:
VT = E - Ir
Finding Current in Simple Circuit
Ohm’s law: Current I is
the ratio of emf E to
total resistance R + r.
VT = IR
VT
r
E
I=
Rr
Cross multiplying gives:
IR + Ir = E; VT = IR
VT = E - Ir
R
I
E
+
r
Battery
-
Example 2. A 3-V battery has an internal
resistance of 0.5 W and is connected to a
load resistance of 4 W. What current is
delivered and what is the terminal
potential difference VT?
E
3V
R
I=

r
R  r 4 W  0.5 W
I = 0.667 A
VT = E – Ir
VT = 3 V – (0.667 A)(0.5 W)
VT = 2.67 V
R=4W
I
E=3V
+
r = 0.5 W
-
Power in Circuits
Recall that the definition of power is work or
energy per unit of time. The following apply:
2
V
P  VI ; P  I R; P 
R
2
The first of these is normally associated with the
power gains and losses through emf’s; the latter
two are more often associated with external loads.
Power, Potential, and EMF
Consider simple circuit:
Terminal
Voltage
VT
Multiply each term by I:
VTI = EI - I2r
r
E
VT = E - Ir
+
I
Battery
-
R
The power delivered to the external circuit is
equal to the power developed in the emf less
the power lost through internal resistance.
Example 3. The 3-V battery in Ex. 2 had
an internal resistance of 0.5 W and a
load resistance of 4 W. Discuss the
power used in the circuit.
From Ex. 2, we found:
I = 0.667 A
R
r
VT = 2.67 V
Power developed in emf:
R=4W
I
E=3V
EI = (3.0 V)(0.667 A) = 2.0 W
Power lost in internal r:
I2r = (0.667 A)2(0.5 W) = 0.222 W
+
r = 0.5 W
-
Example 3 (Cont.). Discuss the power
used in the simple circuit below.
Power in emf: EI = 2.00 W
Power loss:
I2r = 0.222 W
Power lost in external load R:
I2R = (0.667)2(4 W) = 1.78 W
This power can also be found
using VT = 2.67 V
VTI = (2.67)(0.667 A) = 1.78 W
R
r
R=4W
I
E=3V
+
r = 0.5 W
-
Actual power
used externally.
Example 3 (Cont.). Discuss the power
used in the simple circuit below.
Power in emf: EI = 2.00 W
Power loss in
internal r:
R
r
I2r = 0.222 W
R=4W
I
E=3V
Power lost in external load R:
I2R = VTI = 1.78 W
VTI = EI - I2r
+
r = 0.5 W
-
1.78 W = 2.00 W – 0.222 W
A Discharging EMF
When a battery is discharging,
there is a GAIN in energy E as
chemical energy is converted to
electrical energy. At the same
time, energy is LOST through
internal resistance Ir.
A
12 V, 1 W
+
E
r
-
B
I=2A
Discharging
Discharging: VBA = E - Ir
GAIN
LOSS
12 V - (2 A)(1 W) = 12 V - 2 V = 10 V
If VB= 20 V, then VA = 30 V; Net Gain = 10 V
Charging: Reversing Flow Through EMF
When a battery is charged
(current against normal
output), energy is lost through
chemical changes E and also
through internal resistance Ir.
A
12 V, 1 W
+
E
r
-
I=2A
Charging
Charging: VAB = E + Ir
LOSS
LOSS
-12 V - (2 A)(1 W) = -12 V - 2 V = -14 V
If VA= 20 V, then VB = 6.0 V; Net Loss = 14 V
B
Power Gain for Discharging EMF
Recall that electric power is either VI or I2R
When a battery is discharging,
there is a GAIN in power EI as
chemical energy is converted to
electrical energy. At the same
time, power is LOST through
internal resistance I2r.
A
12 V, 1 W
+
E
r
-
B
I=2A
Discharging
Net Power Gain: VBAI = E I- I2r
(12 V)(2 A) - (2 A)2(1 W) = 24 W - 4 W = 20 W
Power Lost on Charging a Battery
Recall that electric power is either VI or I2R
When a battery is charged
(current against normal
output), power is lost through
chemical changes EI and
through internal resistance Ir2.
A
12 V, 1 W
+
E
r
-
B
I=2A
Charging
Net Power Lost= EI + I2r
(12 V)(2 A) + (2 A)2(1 W) = 24 W + 4 W = 24 W
Example 4: A 24-V generator is used to charge
a 12-V battery. For the generator, r1 = 0.4 W
and for the battery r2 = 0.6 W.
The load resistance is 5 W.
12 V .6 W
First find current I:
E
24V  12V
I

 R 5W  0.4W  0.6W
Circuit current: I = 2.00 A
What is the terminal voltage
VG across the generator?
VT = E – Ir = 24 V – (2 A)(0.4 W)
+
E2
r2
-
I
R
5W
.4 W
24 V
+
E1 r1
-
I
VG = 23.2 V
Example 4: Find the terminal
voltage VB across the battery.
Circuit current: I = 2.00 A
12 V
+
E2
VB = E + Ir = 12 V + (2 A)(0.4 W)
Terminal VB = 13.6 V
Note: The terminal voltage across
a device in which the current is
reversed is greater than its emf.
.6 W
r2
-
I
R
5W
.4 W
24 V
+
E1 r1
-
I
For a discharging device, the terminal voltage is
less than the emf because of internal resistance.
Ammeters and Voltmeters
A
+
V
Voltmeter
Emf
Rheostat
-
Source of
EMF
Ammeter
Rheostat
The Ammeter
An ammeter is an instrument used to measure
currents. It is always connected in series and its
resistance must be small (negligible change in I).
A
E
+
-
rg
Digital readout indicates
current in A
Ammeter has
Internal rg
The ammeter draws just enough current Ig to
operate the meter; Vg = Ig rg
Galvanometer: A Simple Ammeter
The galvanometer uses
torque created by small
currents as a means to
indicate electric current.
A current Ig causes the
needle to deflect left or
right. Its resistance is Rg.
20
N
10 0 10
20
S
The sensitivity is determined by the current
required for deflection. (Units are in Amps/div.)
Examples: 5 A/div; 4 mA/div.
Example 5. If 0.05 A causes full-scale
deflection for the galvanometer below, what
is its sensitivity?
0.05A
mA
Sensitivity 
 2.50
20 div
div
Assume Rg = 0.6 W and
that a current causes the
pointer to move to “10.”
What is the voltage drop
across the galvanometer?
2.5 mA
I
10 div   25 mA
div
20
N
10 0 10
20
S
Vg = (25 mA)(0.6 W
Vg = 15 mV
Operation of an Ammeter
The galvanometer is often the working element
of both ammeters and voltmeters.
A shunt resistance in parallel
with the galvanometer allows
most of the current I to bypass the meter. The whole
device must be connected in
series with the main circuit.
Ig
I
Rg
Rs
Is
I = Is + Ig
The current Ig is negligible and only enough to
operate the galvanometer. [ Is >> Ig ]
Shunt Resistance
Ig
Current Ig causes full-scale
deflection of ammeter of
resistance Rg. What Rs is
needed to read current I
+
from battery VB?
VB
Junction rule at A:
-
Ammeter
Is
Rg
A Rs
R
I = 10 A
I = Ig + Is
Or
Is = I - Ig
Voltage rule for Ammeter:
0 = IgRg – IsRs; IsRs = IgRg
(I – Ig)Rs = IgRg
Rs 
I g Rg
I  Ig
Example 6. An ammeter
has an internal resistance
of 5 W and gives fullscale defection for 1 mA.
To read 10 A full scale,
what shunt resistance Rs
is needed? (see figure)
Rs 
I g Rg
I  Ig
(0.001A)(5 W)
Rs 
10 A  (0.001W
Ig
1 mA
+
VB
-
Ammeter
rg
5W
A rg
R
I = 10 A
Rs = 5.0005 x 10-4 W
The shunt draws 99.999% of the external current.
Operation of an Voltmeter
The voltmeter must be connected in parallel
and must have high resistance so as not to
disturb the main circuit.
A multiplier resistance Rm is
added in series with the
galvanometer so that very
little current is drawn from
the main circuit.
The voltage rule gives:
Ig
Rg
Rm
I
VB
VB = IgRg + IgRm
Multiplier Resistance
Voltmeter
Current Ig causes full-scale
deflection of meter whose
resistance is Rg. What Rm is
needed to read voltage VB
of the battery?
VB = IgRg + IgRm
IgRm = VB - IgRg
Rm 
VB  I g Rg
Ig
Rg
Rm
VB
I
R
Which simplifies to:
VB
Rm 
 Rg
Ig
Example 7. A voltmeter
has an internal resistance
of 5 W and gives fullscale deflection for 1 mA.
To read 50 V full scale,
what multiplier resistance
Rm is needed? (see figure)
VB
Rm 
 Rg
Ig
50 V
Rm 
 5W
0.001A
Ig
Voltmeter
Rg
1 mA
5W
Rm
VB
I
R
Rm = 49995 W
The high resistance draws negligible current in meter.
Summary of Formulas:
Discharging: VT = E - Ir
Power: VTI = EI - I2r
+
r
E
-
I
Charging
+
r
E
-
I
Discharging
Charging: VT = E + Ir
Power: VTI = EI + I2r
Summary (Continued)
Ammeter
Voltmeter
Ig
Rg
+
Rm
A Rs
VB
-
Rg
R
I
Rs 
VB
I
I g Rg
I  Ig
VB
Rm 
 Rg
Ig
R
CONCLUSION: Chapter 28B
EMF and Terminal P.D.