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Charge and CurrentCurrent is defined as rate of flow of charge.

It follows that, if charge Q (in coulombs C) passes some point in a circuit in t seconds, the current I (in amperes A) is given by the equation:

I = Q/t

or: Q = It

N.B. The charge in circuits is usually in the form of electrons - which are negative. (i.e. The charge carriers are negative.) By convention, current flows in the opposite direction to the flow of electrons, as shown below:

VoltageVoltage (or potential difference) is defined as the energy transferred to the components in a circuit per unit charge passing through them.

It follows that:

V = W/Q REMEMBER!

where V is the voltage (in volts V) between 2 points in a circuit, Q is the charge (in C) passing from one point to the other, and W is the energy (in J) transferred in the process.

It also follows that: 1 V = 1 JC-1

ResistanceConsider the circuit below:

The resistance R (in ohms ) of the component between A and B is defined as the voltage V (in V) between A and B divided by the current I (in A) flowing through the component.

i.e. R = V/I REMEMBER!

Ohm's LawSuppose the voltage V across a component is varied and the current I is measured - and graphs of V against I are plotted. Typical results are shown below:

A component (or conductor) is said to obey Ohm's Law if the graph is a straight line - like the one above left. In this case I is proportional to V. The conductor is also said to be ohmic.

The resistance (V/I) of an ohmic conductor is constant. The resistance (V/I) of a non-ohmic conductor varies.

Resistors in Series and ParallelResistors in series are illustrated below:

R1 and R2 can be replaced by a single resistor R, where:

R = R1 + R2

If you want to prove the above formula, the starting point is the relationship between the voltages:

   V = V1 + V2

The above relationship is a consequence of conservation of energy.

Resistors in parallel are illustrated below:

R1 and R2 can be replaced by a single resistor R, where:

1/R = 1/R1 + 1/R2

If you want to prove the above formula, the starting point is the relationship between the currents:

   I = I1 + I2

The above relationship is a consequence of conservation of charge.

Electrical Energy and PowerFrom the definition of voltage (see above):

electrical energy transferred to a component = (voltage across it) x (charge passing through it)

i.e. W = VQ

Also:

Charge (Q) = current (I) x time (t)

Therefore, combining the above 2 equations:

electrical energy W = VIt REMEMBER!

Power = (energy transferred)/(time taken)

Therefore, from the above equation:

power P = VIt/t

i.e. P = VI REMEMBER!

Since V = IR, it follows that: P = I2R

Also, since I = V/R, it follows that: P = V2/R

E.M.F. and Internal ResistanceCells, batteries, mains power supplies, etc all have an e.m.f. and internal resistance.

e.m.f. stands for "electromotive force", but it is really a voltage - the maximum voltage that the battery etc can supply.

Internal resistance r is self-explanatory. It is the resistance inside the battery etc.

Consider the circuit below:

The battery etc is enclosed by a circle. It consists of an e.m.f and an internal resistance r.

The terminal p.d. V is the voltage across the terminals of the battery etc - which is generally a bit lower than the e.m.f.

The load resistance R is the resistance connected externally across the terminals of the battery etc.

The e.m.f. drives the current I through both the external resistance R and the internal resistance r.

Therefore: = I(R + r) = IR + Ir

But: IR = V = terminal p.d.

Therefore: = V + Ir

Or: V = - Ir

Thus V is always less than unless either I = 0 or r = 0.

i.e. The terminal p.d. is only equal to the e.m.f. if no current is drawn or the internal resistance equals zero.

Maximum Power TheoremConsider this circuit again:

The power P delivered to the load resistance R is given by:

P = VI = I2R

Theory shows that the power delivered is a maximum when the external resistance

R is equal to the internal resistance r.

i.e. P = max when R = r

The graph below shows how the power peaks when R = r:

Efficiencypercentage efficiency = {useful energy (or power) output}/{total energy (or power) input} x 100%

Suppose we apply the above to a battery connected to a load resistance:

The useful power output = the power delivered to the load = I2R

The total power input = the power delivered to the load + internal resistance = I2(R + r)

Therefore: percentage efficiency = {I2R}/{I2(R + r)} = R/(R + r)

According to the maximum power theorem, maximum power is delivered to the load when R = r.

Efficiency then = r/(r + r) x 100% = 50%

To approach 100% efficiency, R would have to be very large (or r would have to be nearly zero) - but very little power would be delivered in either case.

i.e. In the case of batteries etc, maximum power is not compatible with maximum efficiency.

Resistance and TemperatureThe resistance of a metal increases as the temperature increases. i.e. Metals are said to have positive temperature coefficients of resistance (PTC).

The resistance of a semiconductor (e.g. a thermistor) decreases as the temperature increases. i.e. Semiconductors are said to have negative temperature coefficients of resistance (NTC).

The diagrams below illustrate the reasons for this difference:

In the case of the semiconductor the increase in the number of free electrons more than compensates for the increased lattice vibration - so resistance decreases as temperature increases.

(N.B. The above explanation assumes that the semiconductor is n-type - i.e. that the charge carriers are electrons (negative). Some semicnoductors (p-type) have positive charge carriers (called "holes") - which are, in effect, sites in atoms where electrons are missing.)

Specific Heat CapacityThe specific heat capacity c of a substance is the amount of energy required to raise the temperature of 1 kg of it by 1 oC (or by 1 K).

i.e: c = E/m

..... where E is the amount of energy (in J) transferred to the substance, m is its mass (in kg), and is its temperature rise (in oC).

Thus the units of c are Jkg-1oC-1.

Re-arranging the above equation:

E = mc

Continuous Flow Heat TransferConsider the radiator below:

Suppose that water is pumped through it at a rate of m kgs-1 and that it transfers E J of energy from the hot water to the surroundings in t s. This causes the temperature of the water to fall from 1 at the inlet to 2 at the outlet.

Mass of water flowing through radiator in t s = mt.

Therefore, from the equation in the previous section: E = (mt)c(1 - 2)