Transcript 5.3 Emf and internal resistance

5.3 Emf and internal resistance
Internal resistance
* is the opposition to charge moving through a source of electricity
* causes electrical energy to be dissipated inside the source as charge
moves through it
Producing heat !
Electromotive force of a source
* is the potential (chemical) energy transferred to electrical energy when
one coulomb of charge passes through it
E =W
Q
5.3 Emf and internal resistance
A high resistance voltmeter measures terminal Pd.
E
V
r
Open circuit Pd of a cell equals the EMF in volts,
because the current through the cell is negligible.
Energy supplied
per coulomb
by cell
I
E
Energy wasted
per coulomb
by internal resistance
=
Energy changed
per coulomb
by external circuit
EMF
=
Pd across R
+
Pd across r
E
=
V
+
v
E
=
IR
+
Ir
E
=
I ( R + r)
+
R
r
5.3 Emf and internal resistance
Q. Find the internal resistance of a cell if its emf is 3V
and the pd across an external resistance is 2.5 volts
when 0.5 A flows.
E
V
r
EMF
=
Pd across R
+
Pd across r
E
=
V
+
v
“terminal pd”
I
E
E
=
IR
+
Ir
3
=
2.5
+
0.5 r
0.5
=
0.5 r
r
=
1.0 ohm
R
r
“lost volts”
5.3 Emf and internal resistance
Power
Power supplied by cell
= IE
= I(IR +Ir)
= I 2 R + I 2r
Useful power delivered to R
Maximum power delivered to R
Power
delivered
to R
* occurs when the load resistance R
matches the internal resistance r
I2 R
=
E2
R
(R+r)2
(R+r)I =
I =
E
E
(R+r)
r
Load resistor R
5.3 Emf and internal resistance
Measurement of emf and internal resistance r
* adjust the variable resistor for
different values of current and
record the terminal pd across the cell
r
R limits the
current to
safe levels !
Terminal
Pd /V
2
Emf = 1.5V
1
0.5
1.0
Circuit current / A
5.3 Emf and internal resistance
Measurement of emf and internal resistance r
Terminal
Pd /V
2
EMF
=
Pd across R
+
Pd across r
E
=
V
+
v
E
=
IR
+
Ir
IR
=
E
-
Ir
V
=
E
-
rI
-
mX
Compare with straight line equation:
Y
Emf = 1.5V
=
c
m = gradient
(0, 1.5)
(1.2 , 1)
1
m=
Y
Int resistance r = m = -0.5 = 0.42
0.5
1.0
Circuit current / A
1.2