ELECTRICAL MEASUREMENTS AND METHODS
Fundamentals of electrical measurements:
We are concerned with three quantities - electrical current, electrical potential, and electrical resistance.Electrical current is a measure of the flow of charges per unit time past a point in a circuit. Current is measured in amperes, or amps, and the symbol for current is "I".
1 amp = 1 coulomb/second (an electron has a charge of 1.6 X 10-19 coulomb). Electrical potential V is a measure of the work required to move an electrical charge between two points. If the work W is needed to move charge q from point a to point b, the potential difference between points a and b is
(1)
where "V" is in volts (W in MKS units - newton-meters - and q in coulombs). Each free electron in a conductor is accelerated by the electric field until it collides with an obstruction and loses some energy and velocity. The resistance R to current flow I is related to potential difference V by Ohm's Law
(2)
where V is in volts, I in amperes and R in ohms. If we consider a cylindrical wire conductor of length L and cross-sectional area A, the resistance of that wire is
(3)
where 
is the electrical resistivity of
the material. Note that R is not constant for all conditions. is often a function of temperature, and R can be a function
of current. As the number of electron paths available in a conductor becomes saturated,
electrons themselves begin to collide and the resistance increases. In other words, equation
2 is valid only for a limited range of electrical currents.
Circuit elements: a conventional DC electrical circuit contains batteries, conductors and resistors. For the purpose of circuit analysis, we consider the wires connecting the resistors as having a sufficiently low resistance such that they contribute no significant resistance to the circuit. Figure 1 shows a symbolic circuit diagram with conventional symbols for a battery and resistor.
Figure 1: Symbolic direct current (DC) circuit diagram with battery and resistor.
From Ohm's Law, if (in Figure 1) R = 12 ohms and V = 3 volts, the current I
in the circuit is
I = V/R = 3/12 = 0.25 amps (or 250 milliamps - also written 250 ma).
Series and parallel circuits
Electrical components are in series when current flowing through one component also flows through other components. Electrical components are in parallel when current is divided at a junction and part flows through each component but not through the others. Figure 2 shows a circuit with 4 resistors in series. The resistance of the circuit is equal to the sum of the 4 resistors.
Figure 2: Resistors in series. If R1 = 10 
, R2 = 20 , R3
= 25 and R4 = 6 , the total circuit
resistance = 61 . If the battery voltage is 6
volts, a current of I = 6/61 = 0.098 amps (98 ma). The voltage drop across any one
resistor can be found by applying Ohm's Law: V = I*R = 0.098*10 = 0.98 volts (98
millivolts) across R1, 2.45 volts across R3,
etc.
For "n" resistors in series, the equivalent resistance is equal to the sum of the resistors
(4)
Electrical components in parallel offer less resistance to current than any single component. Figure 3 shows 4 resistors in parallel. The equivalent circuit resistance RT for "n" resistors in parallel is found by the equation
(5)
Figure 3: Resistors are the same as in Figure 2. Now, total resistance (RT) is
(1/RT) = (1/10) + (1/20) + (1/25) + (1/6) = 0.357; 1/0.357 =
2.80 . If the battery has a voltage of 6 volts,
current flowing from the battery is (6/2.80) = 2.14 amps. The voltage drop across the
parallel network (6 volts) is the same across each resistance, and the total current
(I1 + I2 + I3 + I4) = 2.14 amps. The current
through R1 is (6 volts/10 ) = 0.60 amps.
Figure 4 shows a mixed series-parallel circuit. Logical analysis can reduce this
to equivalent total circuit resistance and equivalent resistance of the parallel
components. R1 is in series with the (R2, R3 and R4) network. Using the Figure 2 resistors, the equivalent resistance
of the 3-component parallel network is (1/RT = (1/20) + (1/25) + (1/6) = 0.257; RT = 3.90 + 10 (R1 in series) = 13.9 . For
a 6-volt battery, the current through R1 is I = 6/13.9 = 0.432 amps and
the voltage drop across R1 is V = 10 * .432 = 4.32 volts. The voltage drop
across the parallel network is V = 3.9 * .432 = 1.68 volts; current through R2
is I = 1.68/20 = .084 amps; through R3, 1.68/25 = .067 amps; through R4, I = 1.68/6 =
0.280 amps. Note that (except for round-off error), total voltage drops across circuit
elements equals battery voltage, and, current divided among parallel elements equals the
current entering the network.
Figure 4: combination series-parallel circuit.
Test yourself:
1. In the circuit given (Figure 5), battery voltage is 24 volts. Calculate the voltage drop across each resistor.
2. An electrical current of 100 milliamps passes through a cylinder of water-saturated sand. The diameter of the cylinder is 8 inches and the sand column is 6.0 feet long. The voltage drop across a pair of electrodes 10 cm apart is 245 millivolts. Calculate the resistivity of the water-saturated sand.
3. A voltmeter with input resistance of 100,000 ohms is used to measure the voltage drop across a 100-ohm resistor. Calculate the equivalent resistance of this parallel arrangement.
Figure 5: Circuit for analysis, Test Yourself.