Series Circuits — Two Bulbs on One Battery (Worked Example)

Medium

Problem

Two bulbs with resistances 60 Ω and 120 Ω are connected in series to a 12 V battery. Find (a) the total resistance, (b) the current, and (c) the voltage across each bulb.

Given

R₁ = 60 Ω
R₂ = 120 Ω
V = 12 V
series connection

Find

R_total, current I, V₁ and V₂.

Principle

In series, resistances add and the same current flows through every component; the supply voltage splits across them.

Formula

R_{total} = R_1 + R_2, \quad I = \frac{V}{R_{total}}, \quad V_n = IR_n

R_total = R₁ + R₂; I = V/R_total; V_n = I·R_n

Solution

Total resistance: $R_{total} = 60 + 120 = 180\ \Omega$
Current: $I = 12/180 = 0.0667\ \mathrm{A} = 66.7\ \mathrm{mA}$
Voltage drops: $V_1 = 0.0667 \times 60 = 4.0\ \mathrm{V}$ ; $V_2 = 0.0667 \times 120 = 8.0\ \mathrm{V}$
Check: $4.0 + 8.0 = 12\ \mathrm{V}$ ✓

R = 180 Ω; I ≈ 66.7 mA; V₁ = 4.0 V, V₂ = 8.0 V

Unit check

Ω + Ω = Ω; V/Ω = A; A × Ω = V. All consistent.

What this means physically

The bigger resistor hogs more voltage (8 V vs 4 V) and would glow brighter. This voltage-divider behavior is exactly how dimmer circuits and sensor dividers work.

Common mistake

Assuming each bulb gets the full 12 V. In series the voltage divides — only parallel branches see the full supply voltage.