Circuit Analysis: Energy, Dividers, Kirchhoff's Laws, Bridges, and Op-Amp Gain
Unit 2: Fundamentals of Electricity and Electronics. Lesson 13 of 60 in CBET Academy.
The exam's electronics items are short calculations. Each one below is a formula, a worked example, and the mistake the distractors are built from.
Energy and the kilowatt-hour
Power is the rate of energy use; energy is power times time. A device drawing 500 mA at 120 V uses 0.5 x 120 = 60 W; run for 8 hours it uses 60 x 8 = 480 watt-hours = 0.48 kWh, the unit on the electric bill. P = V x I, P = I squared x R, P = V squared / R: a 10 A current through 30 ohms dissipates 100 x 30 = 3,000 W.
Equivalent resistance
Series adds: 12 + 8 + 6 = 26 ohms. Parallel uses reciprocals: for 4, 8, and 5 ohms, 1/4 + 1/8 + 1/5 = 0.25 + 0.125 + 0.2 = 0.575, so R = 1 / 0.575 = 1.74 ohms, smaller than the smallest branch. For two resistors the shortcut is product over sum.
Voltage divider
In series, each resistor takes a share of the source voltage in proportion to its resistance: V across R1 = V source x R1 / (R1 + R2 + ...). A 10 V source across 1.5 kilohms and 1 kilohm in series puts 10 x 1000 / 2500 = 4 V across the 1 kilohm resistor. Every resistor in a series string obeys the same rule; the shares add back to the source.
Current divider
In parallel, the current splits in inverse proportion to resistance: I through a branch = I total x (1 / R branch) / (sum of 1 / R for all branches). With 600 mA feeding 80, 50, and 200 ohm branches, the 50 ohm branch gets 600 x (1/50) / (1/80 + 1/50 + 1/200) = 600 x 0.02 / 0.0375 = 320 mA. The smallest resistance takes the biggest share. A quicker check: all parallel branches see the same voltage, so I = V / R for each.
Kirchhoff's laws, used properly
- KCL (current law): current into a node equals current out. Nodal analysis: if 9 A enters and 5 A and 2 A leave on two branches, the third branch carries 2 A. A negative answer means the current flows opposite the arrow you assumed.
- KVL (voltage law): the voltages around any closed loop sum to zero; rises equal drops. Mesh analysis: a 10 V source and drops of 3, 2, and 4 V leave 1 V for the last element.
These let you solve any unknown from a schematic when the resistor formulas alone will not.
A battery's internal resistance
A real battery is an ideal source in series with a small resistance. If an open-circuit 1.58 V cell drops to 1.46 V with an 80 ohm load, the load current is 1.46 / 80 = 0.01825 A, and the internal resistance is the lost voltage over that current: (1.58 minus 1.46) / 0.01825 = 6.58 ohms. Internal resistance rises as a battery ages, which is why an old battery reads full voltage unloaded and collapses under load.
The Wheatstone bridge
Four resistors in a diamond with a meter across the middle. The bridge is balanced (zero current in the meter) when the ratios of the two sides match: R1 / R2 = R3 / R4. Given three values, the unknown is R4 = R3 x R2 / R1: with 9.8, 5.8, and 1.9 ohms, the missing resistance is 9.8 / 5.8 x 1.9 = 3.2 ohms. Strain gauge pressure transducers and thermistor circuits are bridges; the tiny imbalance is the signal.
Op-amp gain from resistor ratios
For an inverting amplifier, gain = R feedback / R input (the sign inverts). A gain of 9 with a 5 kilohm input resistor needs a 45 kilohm feedback resistor. Non-inverting gain is 1 + R feedback / R input. ECG and pressure amplifiers are built from these ratios; a drifted resistor is a wrong gain.
Circuit Analysis: Energy, Dividers, Kirchhoff's Laws, Bridges, and Op-Amp Gain, explained slowly
Calculation questions are pattern questions
Every electronics calculation on the exam is one of a handful of patterns. Recognize the pattern, write the formula, substitute, and check the answer against a sanity rule (parallel resistance is smaller than any branch; divider shares add to the source; currents into a node equal currents out). The wrong options are always the result of using the wrong pattern.
Pattern 1: energy from power and time
given formula example V and I P = V x I 120 V x 0.5 A = 60 W I and R P = I squared x R 10 A, 30 ohms: 100 x 30 = 3,000 W = 3 kW V and R P = V squared / R 120 V, 60 ohms: 14,400 / 60 = 240 W P and time E = P x t 60 W x 8 h = 480 Wh = 0.48 kWh
Sanity check: kilowatt-hours are energy; kilowatts are power. Bills are in kWh.
Pattern 2: equivalent resistance
Series: add. Parallel: add reciprocals and invert, or product over sum for two.
4, 8, 5 ohms in parallel: 1/4 + 1/8 + 1/5 = 0.25 + 0.125 + 0.2 = 0.575; R = 1 / 0.575 = 1.74 ohms 120-ohm resistor in series with two 60-ohm resistors in parallel: 60 || 60 = 30; 30 + 120 = 150 ohms
Drawing check: two components in parallel share both end nodes; in series they share exactly one node with nothing else attached.
Pattern 3: dividers
voltage divider (series) V across Rx = V source x Rx / R total 10 V across 1.5 k and 1 k: V across 1 k = 10 x 1000 / 2500 = 4 V 120 V across 60 and 100: V across 60 = 120 x 60 / 160 = 45 V 12 V across 10, 20, and (15 || 30 = 10): total 40; V across the parallel pair = 12 x 10 / 40 = 3 V, the same across the 30 as across the 15 current divider (parallel) I through Rx = I total x (1/Rx) / (sum of 1/R) 600 mA into 80, 50, 200: 1/80 + 1/50 + 1/200 = 0.0125 + 0.02 + 0.005 = 0.0375; I(50) = 600 x 0.02 / 0.0375 = 320 mA 240 V across a 60-ohm branch: I = 240 / 60 = 4 A regardless of the other branches
Sanity: the biggest resistor takes the biggest voltage in series and the smallest current in parallel.
Pattern 4: Kirchhoff
KCL at a node: 9 A in = 5 A + 2 A + I1 out -> I1 = 2 A KCL with a sign: 3 + I2 + 6 = 2 + 3 + 1 -> I2 = -3 A (3 A the other way) KVL in a loop: -10 + 3 + 2 + 4 + V = 0 -> V = 1 V
Nodal analysis uses KCL at nodes; mesh analysis uses KVL around loops. A negative result only means your assumed direction was backward.
Pattern 5: internal resistance
open-circuit voltage 1.58 V; loaded voltage 1.46 V with 80 ohms I = 1.46 / 80 = 0.01825 A r internal = (1.58 - 1.46) / 0.01825 = 6.58 ohms
The lost voltage is dropped inside the battery. A tired battery has high internal resistance: full voltage on the meter, nothing under load.
Pattern 6: the bridge
balanced when R1 / R2 = R3 / R4 (no current through the center meter) R1 = 9.8, R2 = 5.8, R3 = 1.9: R4 = R3 x R2 / R1 = 1.9 x 5.8 / 9.8 = 1.12? careful with which resistor is which: the guide's arrangement gives Rx = (R1 / R2) x R3 = (9.8 / 5.8) x 1.9 = 3.2 ohms
The lesson: write the ratio from the diagram, solve for the unknown, and check that the two side ratios match. Strain gauges and thermistors sit in bridges so a tiny resistance change becomes a measurable voltage.
Pattern 7: op-amp gain
Inverting: gain = R feedback / R input (output inverted). Gain 9 with R in = 5 kilohms means R f = 45 kilohms. Non-inverting: 1 + R f / R in. A drifted feedback resistor changes the gain, which shows up as a wrong calibration.
How the exam asks it
- "A bulb draws 500 mA at 120 V for 8 hours; energy used?" 0.48 kWh.
- "Power with 10 A through 30 ohms?" 3 kW.
- "Voltage across the 60-ohm resistor in series with 100 ohms on 120 V?" 45 V.
- "Current through the 50-ohm branch of 80, 50, 200 with 600 mA?" 320 mA.
- "I1 when 9 A enters and 5 A and 2 A leave?" 2 A.
- "Internal resistance from 1.58 V dropping to 1.46 V into 80 ohms?" 6.58 ohms.
- "R2 for gain 9 with R1 = 5 k?" 45 k.
What to memorize
- P = VI = I squared R = V squared / R; E = P x t; kWh.
- Series adds; parallel reciprocals; two in parallel: product over sum.
- Voltage divider proportional to R; current divider inversely proportional to R.
- KCL: in equals out; KVL: loop sums to zero.
- Internal resistance = voltage drop / load current.
- Bridge balance: equal ratios. Inverting gain = Rf / Rin.