Electromotive force is the energy converted into electrical energy when 1 C of charge passes through the power source
P.D. and E.M.F.
Internal Resistance
Internal resistance is the resistance to current flow within the power source; it reduced p.d. when delivering current
V = Ir – E
Voltage across the resistor: V = IR
Voltage lost to internal resistance: V = Ir
Thus e.m.f.: E = IR + Ir
E = I(R + r)
Kirchhoff’s First Law
Sum of currents into a junction is equal to the sum of currents out of the junction
Kirchhoff’s first law is another statement of the law of conservation of charge
Kirchhoff’s Second Law
Sum of e.m.fs in a closed circuit is equal to the sum of the potential differences
Kirchhoff’s second law is another statement of the law of conservation of energy
Applying Kirchhoff’s Laws
Calculate the current in each of the resistors
Using Kirchhoff’s first law:
I_{3} = I_{1} + I_{2}
Using Kirchhoff’s second law on loop ABEF:
3 = 30I_{3} + 10I_{1}
Using Kirchhoff’s second law on loop CBED:
2 = 30I_{3}
Using Kirchhoff’s second on loop ACDF:
3 – 2 = 10I_{1}
Solve the simultaneous equations:
I_{1} = 0.1
I_{2} = -0.033
I_{3} = 0.067
Deriving Effective Resistance in Series
From Kirchhoff’s second law:
E = ∑IR
IR = IR_{1} + IR_{2}
Current is constant, therefore:
R = R_{1} + R_{2}
Deriving Effective Resistance in Parallel
From Kirchhoff’s first law:
I = ∑I
I = I_{1} + I_{2}
^{V}/_{R} = ^{V}/_{R1} + ^{V}/_{R2}
Voltage is constant, therefore;
^{1}/_{R} = ^{1}/_{R1} + ^{1}/_{R2}
Properties of Magnets
A potential divider divides the voltage into smaller parts
^{Vout}/_{Vin} = ^{R2}/_{RTotal}
Usage of a thermistor at R_{1}:
Resistance decreases with increasing temperature
It can be used in potential divider circuits to monitor and control temperatures
Usage of an LDR at R_{1}:
Resistance decreases with increasing light intensity
It can be used in potential divider circuits to monitor light intensity
Potentiometers
A potentiometer is a continuously variable potential divider used to compare potential differences
Potential difference along the wire is proportional to the length of the wire
It can be used to determine the unknown e.m.f. of a cell
This can be done by moving the sliding contact along the wire until it finds the null point that the galvanometer shows a zero reading; the potentiometer is balanced
For example:
E_{1} is 10 V, distance XY is equal to 1 m. The potentiometer is balanced at point T which is 0.4 m from X. Calculate E_{2}