Transcript Emf and Internal Resistance

Notes p.11
Emf and Internal Resistance
To do
Work in groups of 4 or 5 to carry out Activity 2 –
Internal Resistance.
Internal Resistance
In theory… V = 12 V
In practice… V = 11.8 V
Why???
V
Reminder …. Voltage = Energy per Coulomb
When current flows, wires/battery etc heat up hence
energy is “lost” per coulomb. (So we have some “lost
volts” before the charge even reaches the load
resistor.)
The e.m.f. is the energy supplied per coulomb
BEFORE any volts are lost.
To measure e.m.f. we must have no current flowing
so that there is no energy loss.
battery
R=
V
In the example above we “lost” 0.2 V. We say that
the battery had an “internal resistance” which used
up the 0.2 V. Our circuit can be illustrated as
- terminal
+ terminal
r
e.m.f. = 12 V
R
100W
V
= 11.8 V
1. The voltage (V) measured across the battery
terminals = voltage across the load, R.
It is called the terminal potential difference,
t.p.d..
2. Inside the battery, effectively, we have a supply
e.m.f. of 12 V and a small internal resistance, “r”.
3. To measure e.m.f. remove the load, R, leaving an
air gap so no current flows.
r
e.m.f. = 12 V
100W
12V
So V = e.m.f.
= 12 V
4. To calculate “r”, replace the load resistor, R, then
1st find I :
2nd find r
R = 100 W
V = 11.8 V
I=?
: lost volts = 12 – 11.8
= 0.2 V
I = 0.118 A
r=?
I = V/R
= 11.8 / 100
= 0.118 A
r = lost volts
I
= 0.2 / 0.118
= 1.69 W
Worked Examples (see handout)
Example 1
Example 2 : The voltage across a cell was recorded for various values of
current, using the following circuit.
V
r
A
R
A graph of the results was plotted:
Voltmeter
reading (V)
1.6
1.2
0.8
0.4
0
0.25
0.50
0.75
1.00
Current (A)
Voltmeter
reading
1.6
(V)
1.2
0.8
0.4
0
0.25 0.50 0.75 1.00 Current (A)
a) Determine the e.m.f. of the cell.
2 V (extrapolate line back to when I = 0 A)
b) Calculate the internal resistance of the cell.
PICK A POINT AND WRITE DOWN ALL YOU CAN!
emf = 2V
lost volts = 2 – 0.4 = 1.6V
I = 1A
r = ?
r = lost volts / I
=
=
1.6
/ 1
1.6 W
c) Determine the short circuit current for this circuit.
This means all volts (e.m.f.) used in r!
The battery’s “r” is the only resistance in the circuit.
e.m.f. = 2 V
I=?
r= W
I short circuit = e.m.f.
r
=2
1.6
= 1.25 A
Summary
1. E.m.f is measured when no current flows.
e.m.f. = I RT = I (R + r)
2. t.p.d. is voltage across the load resistance, R:
t.p.d. = e.m.f. – lost volts
t.p.d. = IR
3. “lost volts” is the potential difference across the
internal resistance of the cell.
Lost volts = I r
4. Short circuit current …
Ishort = emf / r
Summary (cont.)
5. Ideal Electrical Supplies:
The maximum power is transferred from a
battery when the
internal resistance = load resistance
Over to you – continue problems up to
p.13-18, Q. 6 - 18
Problems on Circuits
11.
a) 6W
b) 3W
d) 6V
e) 24W
12. a) 11.5A
page 60-61, Q 11-24
c) 2 A for A1, 1.5A for A2
b) 12.8A (12W), 9.6A (8W), 3.2A (24W)
13. a) R1 = 230W, R2 = 115W
b) LOW = 230W, HIGH = 690W
14.a) 4W
b) 36W
15.a) 2V b) 1.6V c) r = 0.5W, R = 2W d) 1.3A, 1.3V
16. 10W
17.
5W
18. 12W
19.a) 1.3V b) current decreases as total resistance
increases. c) tpd increases - as R increases it takes a
larger share of the voltage OR as current increases
there are more “lost volts” in the internal resistance of
the cell.
Problems on Circuits
16. 10W
17.
page 60-61, Q 11-24
5W
18. 12W
19.a) 1.3V b) current decreases as total resistance
increases. c) tpd increases - as R increases it takes a
larger share of the voltage OR as current increases
there are more “lost volts” in the internal resistance of
the cell.
20. 0.3A
21. a) 3W b) I = 0.67A, R = 3.73W
22. 4V
23. a) standard circuit (see notes)
b) (i) 1.1V
(ii) approx 4.10 W
c) 4.00W, 4.00W, 4.17W, 4.00W, 4.10W, 4.17 W.
d) (4.10 ± 0.02 ) W
24. a) 6V
b) 0.1W
c) 60A d) (i) 5.63V (ii) 21.1W