SME
Grade 8 Science · Problem type
More bulbs in series
Use this for “A third identical bulb is added in series. What happens to the current and the brightness?”
Diagram
I = V ÷ (n × R) = 6 V ÷ (3 × 2 Ω)
= 1 A
This is an example. Type your own values to start a new problem.
I (A)
=
V (V)
÷
(
n
×
R (Ω)
)
Formulas
I
=
V
÷
(
n
×
R
)
(current
=
battery
voltage
÷
(bulbs
in
the
loop
×
each
bulb’s
resistance))
1 = 6 ÷ (3 × 2)
Assumptions
•
The bulbs are all the same, in one loop.
•
Each bulb adds its resistance, so more bulbs share less current and each glows dimmer.
•
Bulbs glow by the power they get, so the picture dims as you add bulbs.
Step-by-step
WE KNOW
V = 6 V (battery voltage), n = 3 (bulbs in the loop), R = 2 Ω (each bulb’s resistance)
FIND
current (I)
Step 1 · Find current (I)
Use I = V ÷ (n × R) (current = battery voltage ÷ (bulbs in the loop × each bulb’s resistance))
The
total
resistance
is
n
×
R
;
divide
the
voltage
by
it.
I
=
6
3 × 2
I
=
6
6
I
=
1
A
Check
Put the numbers back in. Both sides should match.
1 = 6 ÷ (3 × 2) ✓
Topic check
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