3. 1/R = 1/20 + 1/20=1/10
R = 10 Ω
Effective R = 10 + 10 + 5 =25 Ω
4. Important!!
• A circuit is a closed loop through which
current can continuously flow.
• The resistance within the cell or battery itself
is called internal resistance.
• a cell consists of electrodes in a chemical
electrolyte. When the cell is connected in a
circuit, the current flowing in the electrolyte
through electrodes experience internal
resistance.
6. ELECTROMOTIVE FORCE (e.m.f)
• The total electrical energy given to one
coulomb of charge flowing through the cell.
• Work done by a source to move a unit of
electrical charge from one terminal to the
other through the complete circuit.
• It is not a force but is a quantity that measures
energy per unit charge.
• Make a current flow through the circuit.
7.
8. e.m.f & potential difference, V
• E.m.f is the work done by a source in driving 1
coulomb of change around a complete circuit.
• The lost volts is the potential difference
required to drive the current through the
internal resistance.
• Terminal potential difference, V is the work
done by a source in driving 1 coulomb of
change through the external resistor.
9. • Electromotive force, E = I (R + r)
• (R+ r) is the total resistance of the circuit.
• V = IR, where R is the sum of external
resistance.
10. • Electromotive force, e.m.f., E = Potential
Difference + Drop in Potential Difference
across resistor, R due to internal resistance,r
= VR + Vr, where VR = IR and Vr = Ir
= IR + Ir
= I (R + r)
11. Example:
Why is the potential difference across the
resistor not the same as the e.m.f. of the
battery?
12. Determine the value of the internal
resistance
• Since E = V + Ir
• 1.5 = 1.1 + 0.5 r
• r = 0.8 Ω
13. • A voltmeter connected directly across a
battery gives a reading of 1.5 V. The voltmeter
reading drops to 1.35 V when a bulb is
connected to the battery and the ammeter
reading is 0.3 A. Find the internal resistance of
the battery.
14. • E = 3.0 V, V = 1.35 V, I = 0.3 A
• Substitute in : E = V + Ir
• 1.5 = 1.35 + 0.3(r)
• r = 0.5 Ω
15. A cell of e.m.f., E and internal resistor, r is
connected to a rheostat. The ammeter reading, I
and the voltmeter reading, V are recorded for
different resistance, R of the rheostat. The graph
of V against I is as shown.
• From the graph, determine
• a) the electromotive force, e.m.f., E
•
• b) the internal resistor, r of the cell
•
19. Electrical Energy and Electrical Power
• Potential difference, V across two points is
the energy,E dissipated or transferred by a
coulomb of charge, Q that moves across the
two points.
20. • Power is defined as the rate of energy
dissipated or transferred.
21.
22.
23.
24. • An electric kettle is rated 240 V 2 kW.
Calculate the resistance of its heating element
and the current at normal usage.
P = IV
I = P/V
= 2000 / 240
= 8.3 A
25. • An electric kettle operates at 240 V and carries
current of 1.5 A.
• (a) How much charge will flow through the
heating coil in 2 minutes
• Q = I t
• = (1.5) (2 x 60)
• = 180 C
