Circuit analysis, measurement and batteries

Lesson 5 introduced current, resistance, Ohm's law, series and parallel resistors and Kirchhoff's laws. With those tools any resistive circuit can be solved, but larger circuits need a system, and practical work raises new questions. What does a sensor connected to a long cable actually deliver to the instrument reading it? Does connecting a meter change the very voltage it is measuring? How do engineers measure a strain of a few millionths with a resistor? How long will a battery run a device, and why does its voltage sag under load? This lesson develops nodal analysis for systematic solution, the Thévenin and Norton equivalents that reduce any network to a simple source, the maximum power transfer theorem, the principles and pitfalls of meters, the Wheatstone bridge and strain gauge, and the practical behaviour of batteries. ELE-202 and ENCM courses take circuit analysis much further.

Nodal analysis

For circuits with more than a few loops, applying Kirchhoff's laws ad hoc becomes error-prone. Nodal analysis organizes the work. Choose one node as the reference (ground, 0 V) and label the voltages of the other nodes as unknowns. At each unknown node, write Kirchhoff's current law: the sum of currents leaving through all connected branches is zero, with each resistor current written as (voltage difference)/R from Ohm's law. The result is one linear equation per node, a system that MATH-125 methods solve by hand for two or three nodes and that computers solve for millions in circuit simulators.

Thévenin and Norton equivalents

Seen from any two terminals, a linear network of sources and resistors behaves exactly like a single voltage source \(V_{Th}\) in series with a single resistance \(R_{Th}\) (Thévenin's theorem), or equivalently a current source \(I_N = V_{Th}/R_{Th}\) in parallel with \(R_{Th}\) (Norton's theorem). This is enormously useful: a complicated sensor circuit, a battery pack or an amplifier output can be summarized by two numbers, and then any load connected to it is a simple series circuit.

To find the equivalent: \(V_{Th}\) is the open-circuit voltage at the terminals; \(R_{Th}\) is the resistance seen looking back into the terminals with the independent sources turned off (voltage sources replaced by short circuits, current sources by open circuits), or equivalently \(V_{Th}\) divided by the short-circuit current. A network can also be reduced step by step, replacing each source and its resistors by a Thévenin equivalent and combining.

import numpy as np

# Network: 12 V source with 2 ohm to node 1; 4 ohm node 1 to ground; 3 ohm node 1 to node 2;
# 6 ohm node 2 to ground; 1 ohm node 2 to node 3 (output terminal); load RL from node 3 to ground.
Vs = 12.0

def solve(RL):
    G = np.zeros((3, 3)); I = np.zeros(3)
    def add(a, b, R):                      # conductance between nodes a and b (-1 = ground)
        g = 1 / R
        for n in (a, b):
            if n >= 0: G[n, n] += g
        if a >= 0 and b >= 0:
            G[a, b] -= g; G[b, a] -= g
    add(0, -1, 2.0); I[0] += Vs / 2.0       # source as a Norton equivalent: 12 V behind 2 ohm
    add(0, -1, 4.0); add(0, 1, 3.0); add(1, -1, 6.0); add(1, 2, 1.0)
    if RL is not None:
        add(2, -1, RL)
    else:
        G[2, 2] += 1e-12                    # open circuit at the output
    return np.linalg.solve(G, I)

Voc = solve(None)[2]
Isc = solve(1e-9)[2] / 1e-9
Rth = Voc / Isc
print(f"Thevenin equivalent at the output: Vth = {Voc:.3f} V, Rth = {Rth:.3f} ohm")
for RL in [1.0, Rth, 10.0]:
    V = solve(RL)[2]
    print(f"Load {RL:5.2f} ohm: voltage {V:.3f} V, power {V**2 / RL:.3f} W")

Thevenin equivalent at the output: Vth = 4.645 V, Rth = 3.516 ohm
Load  1.00 ohm: voltage 1.029 V, power 1.058 W
Load  3.52 ohm: voltage 2.323 V, power 1.534 W
Load 10.00 ohm: voltage 3.437 V, power 1.181 W

Maximum power transfer

A source with internal resistance \(R_{Th}\) delivers power \(P = V_{Th}^2R_L/(R_{Th} + R_L)^2\) to a load \(R_L\). This is zero for a short circuit (no voltage) and for an open circuit (no current), and greatest in between. Setting \(dP/dR_L = 0\) gives the maximum power transfer theorem: the load receives the most power when \(R_L = R_{Th}\), and that power is \(V_{Th}^2/(4R_{Th})\). At this point, half the power is wasted in the source, so the efficiency is only 50%.

The code above confirms that the 3.52 Ω load draws more power (1.53 W) than either 1 Ω or 10 Ω. Matching is the right goal when signals are weak and every microwatt counts: antennas, radio receivers, audio amplifiers driving speakers, and the transmission lines of later courses are matched to their loads. It is the wrong goal for power systems, where efficiency matters most: a utility generator or a battery is operated with a load resistance far larger than its internal resistance, delivering power at high efficiency well below the theoretical maximum.

Meters and loading

An ammeter measures current, so it is placed in series with the branch, and it should have a very small resistance so that it does not change the current. A voltmeter measures the potential difference between two points, so it is connected in parallel, and it should have a very large resistance so that it draws negligible current. Modern digital multimeters typically have an input resistance of about 10 MΩ on voltage ranges. That is usually enough, but not always: connecting any meter changes the circuit it measures, an effect called loading.

Ammeters must never be connected across a voltage source: their low resistance makes a near short circuit, which blows the meter's fuse or worse. Clamp meters avoid breaking the circuit altogether by sensing the magnetic field around a conductor (Lesson 7), and are the standard tool for measuring currents in electrical installations.

The Wheatstone bridge and strain gauges

Some sensors change their resistance by tiny amounts. A strain gauge, a thin foil grid bonded to a structure, changes resistance in proportion to the strain it experiences: \(\Delta R/R = GF \cdot \varepsilon\), with a gauge factor GF of about 2 for metal foil gauges. A strain of 500 microstrain (5 × 10⁻⁴), a typical working strain in steel, changes a 350 Ω gauge by only 0.35 Ω. Measuring such a small change directly is hopeless; the Wheatstone bridge makes it easy.

A bridge has four resistors in a diamond, with a supply across one diagonal and a voltmeter across the other. When the ratios of the two halves are equal, the bridge is balanced and the output is zero; a small change in one arm produces a small output voltage that can be amplified, starting from zero rather than from a large background voltage. With one active gauge (a quarter bridge) and supply \(V_s\), the output is approximately

Batteries in practice

A real battery behaves approximately as an ideal emf in series with an internal resistance (Lesson 5), so its terminal voltage sags as current increases and recovers when the load is removed. Its capacity is stated in ampere-hours (Ah): a 100 Ah battery can in principle supply 5 A for 20 hours. The energy it stores is capacity times average voltage, in watt-hours. Currents are often expressed as a C-rate: 1C discharges the battery in one hour, 0.5C in two.

Real capacity depends on how the battery is used. Lead-acid batteries deliver noticeably less than their rated capacity at high currents (Peukert's effect) and should not be discharged below about half capacity if they are to last; lithium-ion batteries hold their capacity better at high rates and tolerate deeper discharge, but lose capacity in the cold, a significant effect in Canadian winters and northern US states. Capacity also fades over hundreds to thousands of charge cycles.