7.1 Ohm's Law — Series & Parallel Circuits

7.1 Ohm's Law — Series & Parallel Circuits

Quick Answer: V = IR. Series: current is the same everywhere; voltages add; R_total = R1 + R2 + …. Parallel: voltage is the same across every branch; currents add; 1/Req = 1/R1 + 1/R2 + … (two resistors: Req = R1 R2 / (R1 + R2)). A voltage divider yields Vout = Vin × (R_lower / R_total).

Element 3 Topic 3-B (key topics roughly 009–018) is the quantitative core of general radiotelephone theory. Before you can size loads, bias stages, or reason about RF networks, you must compute resistive circuits the way a bench tech does—with Ohm’s Law and series/parallel combination rules that never change.

Ohm’s Law in three forms

Georg Ohm’s relationship for a pure resistance (or the resistive part of a network) is:

[ V = I R \qquad I = \frac{V}{R} \qquad R = \frac{V}{I} ]

Units: volts (V), amperes (A), ohms (Ω). Keep prefixes consistent: milliamps with kilohms often cancel neatly ((I,\mathrm{mA} = V / R_{\mathrm{k}\Omega})).

Worked example — classic pool arithmetic

What current flows through a 50 Ω resistor with 100 V applied?

[ I = \frac{V}{R} = \frac{100,\mathrm{V}}{50,\Omega} = 2,\mathrm{A} ]

Trap answers on Element 3 style stems include 0.5 A (inverted formula), 5 A (misread 100/20), or absurd 5000 A (unit mix-ups). Always write the formula first, then substitute.

Worked example — finding R from meter readings

A resistor drops 12 V while carrying 40 mA. What is R?

[ R = \frac{12}{0.040} = 300,\Omega ]

Convert milliamps to amperes before dividing, or use (R = 12 / 40,\mathrm{mA} = 0.3,\mathrm{k}\Omega = 300,\Omega).

Series circuits — same current, voltages add

In a series connection there is one path for current. Therefore:

1. Current is the same through every series element.
2. Voltage drops add to equal the source: (V_s = V_1 + V_2 + \cdots).
3. Total resistance is the sum: (R_T = R_1 + R_2 + \cdots).
4. Each voltage drop is (V_n = I R_n) with the same I.

Worked example — three series resistors

A 24 V supply feeds R1 = 100 Ω, R2 = 200 Ω, and R3 = 300 Ω in series.

1. (R_T = 100 + 200 + 300 = 600,\Omega)
2. (I = 24 / 600 = 0.040,\mathrm{A} = 40,\mathrm{mA})
3. Drops: (V_1 = 4,\mathrm{V}), (V_2 = 8,\mathrm{V}), (V_3 = 12,\mathrm{V}) (sum = 24 V)

Notice the voltage shares the same proportions as the resistances (100:200:300 = 1:2:3). That proportion is exactly the voltage-divider idea below.

Parallel circuits — same voltage, currents add

In a parallel connection every branch is connected across the same two nodes. Therefore:

1. Voltage is the same across every parallel branch.
2. Branch currents add to the total source current: (I_T = I_1 + I_2 + \cdots).
3. Equivalent resistance satisfies:

[ \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \cdots ]

For two resistors:

[ R_{eq} = \frac{R_1 R_2}{R_1 + R_2} ]

Critical exam fact: (R_{eq}) of a parallel group is always less than the smallest individual resistor in that group. If someone offers an answer larger than the smallest R, reject it.

Equal resistors shortcut

n identical resistors of value R in parallel:

[ R_{eq} = \frac{R}{n} ]

Three 30 Ω resistors in parallel:

[ R_{eq} = \frac{30}{3} = 10,\Omega ]

That exact pattern appears in Element 3-style practice: equal values → divide by count. Trap options often list the series sum (90 Ω) or a product (270 Ω).

Worked example — two unequal parallels

R1 = 60 Ω and R2 = 40 Ω in parallel across 12 V:

1. (R_{eq} = (60 \times 40)/(60 + 40) = 2400/100 = 24,\Omega)
2. (I_T = 12 / 24 = 0.5,\mathrm{A})
3. Branch currents: (I_1 = 12/60 = 0.2,\mathrm{A}), (I_2 = 12/40 = 0.3,\mathrm{A}) (sum 0.5 A)

Series–parallel combinations

Real schematics mix both topologies. Method:

1. Collapse pure series groups into one equivalent.
2. Collapse pure parallel groups into one equivalent.
3. Repeat until one equivalent resistance remains; then apply Ohm’s Law to find total current; finally back-substitute for branch currents and drops.

Worked example — series with a parallel pair

A 48 V source feeds R1 = 20 Ω in series with the parallel combination of R2 = 60 Ω and R3 = 30 Ω.

1. Parallel pair: (R_{23} = (60 \times 30)/(60 + 30) = 20,\Omega)
2. Total: (R_T = 20 + 20 = 40,\Omega)
3. (I_T = 48/40 = 1.2,\mathrm{A})
4. Drop on R1: (V_1 = 1.2 \times 20 = 24,\mathrm{V}); remaining 24 V appears across the parallel pair.
5. Currents: (I_2 = 24/60 = 0.4,\mathrm{A}), (I_3 = 24/30 = 0.8,\mathrm{A}).

Voltage dividers

A string of series resistors across a source creates intermediate voltages. For a two-resistor divider with R_top toward the high side and R_bottom toward ground (unloaded):

[ V_{out} = V_{in} \times \frac{R_{bottom}}{R_{top} + R_{bottom}} ]

Worked example — unloaded divider

(V_{in} = 15,\mathrm{V}), (R_{top} = 2,\mathrm{k}\Omega), (R_{bottom} = 1,\mathrm{k}\Omega):

[ V_{out} = 15 \times \frac{1}{2+1} = 5,\mathrm{V} ]

Loading warning: if a low-resistance load is placed across (R_{bottom}), the parallel combination lowers the bottom resistance and pulls Vout down. For exam estimation, either assume “no load” as the stem states, or recompute the parallel of (R_{bottom}) and (R_{load}) first.

Current dividers (parallel dual)

Two resistors in parallel share current inversely with resistance:

[ I_1 = I_T \times \frac{R_2}{R_1 + R_2} ]

The smaller resistor takes the larger share of current—same physics as Ohm’s Law at equal voltage.

Power in resistive networks (preview of §7.3)

Once you know V and I (or I and R, or V and R), local dissipation is:

[ P = VI = I^2 R = \frac{V^2}{R} ]

In series, the largest resistor dissipates the most power (same I, larger R). In parallel, the smallest resistor dissipates the most power (same V, smaller R → larger I and larger (V^2/R)).

Exam-day checklist for §7.1

1. Write V = IR and rearrange only after labeling knowns.
2. Ask: series or parallel? Same I vs same V is the fork.
3. Series total R = sum; parallel equal R’s → R/n; two unequal → product/sum.
4. Parallel Req must be < smallest R.
5. Voltage divider: fraction of total R on the bottom leg (unloaded).
6. Back-check: series voltages sum to source; parallel currents sum to total.

Master these six checks and you clear the Ohm’s-Law and network slice of Topic 3-B before moving into frequency, wavelength, and waveform amplitude math.