Students will be able to:
Vₙ = I Rₙ and Vtotal = V₁ + V₂ + V₃ + …
Etotal = E₁ + E₂ + …
(internal resistances add in the same way as ordinary resistors).
Vbranch = Vsource.
\displaystyle\frac{1}{R_{\text{total}}}= \frac{1}{R₁}+ \frac{1}{R₂}+ \frac{1}{R₃}+ …
I_{\text{total}} = I₁ + I₂ + I₃ + ….
In a steady‑state circuit charge cannot build up at any point. If more charge entered a segment than left it, an increasing electric field would develop, opposing further flow. Because a series circuit has no branching points, the same amount of charge that leaves the source per second must pass through every component, giving a single, uniform current I throughout.
All branches connect the same two nodes. The electric potential difference between those nodes is fixed by the source, so each branch experiences the same p.d. Even though the resistances differ, the voltage across each branch remains equal to the source voltage.
| Feature | Series | Parallel |
|---|---|---|
| Path for current | Single continuous path | Multiple independent paths |
| Current through each element | Same everywhere: I_{\text{total}} = I₁ = I₂ = … | Different in each branch: I_{\text{total}} = I₁ + I₂ + … |
| Voltage across each element | Divides according to resistance: V_{\text{total}} = V₁ + V₂ + … | Same across every branch: V{\text{branch}} = V{\text{source}} |
| Total resistance | R_{\text{total}} = R₁ + R₂ + … (adds) | \displaystyle\frac{1}{R_{\text{total}}}= \sum\frac{1}{Rₙ} (decreases) |
| Combined e.m.f. of cells | E_{\text{total}} = E₁ + E₂ + … | Same as series – cells in parallel give the same e.m.f. but the capacity (Ah) adds. |
Problem: A 12 V battery supplies three resistors in series: R₁ = 2 Ω, R₂ = 3 Ω, R₃ = 5 Ω. Find the current through the circuit and the voltage across each resistor.
R_{\text{total}} = 2 Ω + 3 Ω + 5 Ω = 10 Ω
I = V{\text{total}} / R{\text{total}} = 12 V / 10 Ω = 1.2 A
V₁ = I R₁ = 1.2 A × 2 Ω = 2.4 V
V₂ = I R₂ = 1.2 A × 3 Ω = 3.6 V
V₃ = I R₃ = 1.2 A × 5 Ω = 6.0 V
Problem: A 9 V battery feeds three resistors in parallel: R₁ = 3 Ω, R₂ = 6 Ω, R₃ = 9 Ω. Determine (a) the total resistance, (b) the total current drawn, and (c) the current through each resistor.
\[
\frac{1}{R_{\text{total}}}= \frac{1}{3}+\frac{1}{6}+\frac{1}{9}= \frac{6+3+2}{18}= \frac{11}{18}
\]
R_{\text{total}} = \dfrac{18}{11}\;Ω \approx 1.64 Ω
I{\text{total}} = V{\text{source}} / R_{\text{total}} = 9 V / 1.64 Ω \approx 5.5 A
I₁ = 9 V / 3 Ω = 3.0 A
I₂ = 9 V / 6 Ω = 1.5 A
I₃ = 9 V / 9 Ω = 1.0 A
Verify: 3.0 A + 1.5 A + 1.0 A = 5.5 A = I_{\text{total}}.
Problem: Two identical 1.5 V dry cells are connected in series with a 3 Ω resistor. Find the current supplied by the cells (ignore internal resistance).
E_{\text{total}} = 1.5 V + 1.5 V = 3.0 V
R_{\text{total}} = 3 Ω
I = E{\text{total}} / R{\text{total}} = 3.0 V / 3 Ω = 1.0 A.
Two resistors R₁ and R₂ in series can be used to obtain a fraction of the source voltage:
\[
V{\text{out}} = V{\text{source}} \frac{R₂}{R₁+R₂}
\]
(where Vout is measured across R₂).
At any node, the algebraic sum of currents is zero:
\[
\sum I{\text{in}} = \sum I{\text{out}}
\]
This is simply the statement that charge does not accumulate at a junction.
A circuit is powered by a 15 V battery. It contains two parallel branches. Branch A has a single resistor RA = 5 Ω. Branch B consists of three resistors in series: R{B1}=2 Ω, R{B2}=3 Ω, R{B3}=6 Ω. Determine:
Solution outline:
\[
\frac{1}{R_{\text{eq}}}= \frac{1}{5} + \frac{1}{11}= \frac{11+5}{55}= \frac{16}{55}
\]
\[
R_{\text{eq}} = \frac{55}{16}\;Ω \approx 3.44\;Ω
\]
I_A = 15 V / 5 Ω = 3.0 A
I_B = 15 V / 11 Ω ≈ 1.36 A
(Check: 3.0 A + 1.36 A ≈ 4.36 A).
I{B1} = I{B2} = I_{B3} = 1.36 A.
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