SME
Grade 8 Science · Problem type
Bulbs in parallel
Use this for “Two bulbs are wired in parallel across a 6 V battery. What current does each get, and what does the meter read?”
Diagram
Each branch gets the whole battery: I = V ÷ R.
I₁: 6 V ÷ 2 Ω = 3 A
I₂: 6 V ÷ 4 Ω = 1.5 A
I = 3 A + 1.5 A = 4.5 A
This is an example. Type your own values to start a new problem.
I₁ (A)
=
V (V)
÷
R₁ (Ω)
I₂ (A)
=
V (V)
÷
R₂ (Ω)
I (A)
=
I₁ (A)
+
I₂ (A)
Formulas
I₁
=
V
÷
R₁
(first
branch
current
=
battery
voltage
÷
first
bulb’s
resistance)
3 = 6 ÷ 2
I₂
=
V
÷
R₂
(second
branch
current
=
battery
voltage
÷
second
bulb’s
resistance)
1.5 = 6 ÷ 4
I
=
I₁
+
I₂
(current
at
the
meter
=
first
branch
current
+
second
branch
current)
4.5 = 3 + 1.5
Assumptions
•
Each bulb has its own branch across the battery, so each gets the full voltage.
•
The branch currents add at the meter: more branches, more current from the battery.
•
One bulb burning out does not put the other out. That is how house wiring works.
Step-by-step
WE KNOW
V = 6 V (battery voltage), R₁ = 2 Ω (first bulb’s resistance), R₂ = 4 Ω (second bulb’s resistance)
FIND
first branch current (I₁), second branch current (I₂), current at the meter (I)
Step 1 · Find first branch current (I₁)
Use I₁ = V ÷ R₁ (first branch current = battery voltage ÷ first bulb’s resistance)
The branch gets the whole voltage: divide it by the resistance.
I₁
=
6
2
I₁
=
3
A
Step 2 · Find second branch current (I₂)
Use I₂ = V ÷ R₂ (second branch current = battery voltage ÷ second bulb’s resistance)
The branch gets the whole voltage: divide it by the resistance.
I₂
=
6
4
I₂
=
1.5
A
Step 3 · Find current at the meter (I)
Use I = I₁ + I₂ (current at the meter = first branch current + second branch current)
The branch currents join at the meter.
I
=
3
+
1.5
I
=
4.5
A
Check
Put the numbers back in. Both sides should match.
3 = 6 ÷ 2 ✓
1.5 = 6 ÷ 4 ✓
4.5 = 3 + 1.5 ✓
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