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Grade 8 Science · Physical Science

Electric charge, current and simple circuits

Use this for “A 6 V battery runs a 2 Ω and a 4 Ω resistor in series. What is the current?”

Diagram

+−6 VclosedR₁ = 2 ΩR₂ = 4 Ω02AI = 1 A
I = V ÷ (R₁ + R₂)
= 6 V ÷ (2 Ω + 4 Ω) = 1 A
This is an example. Type your own values to start a new problem.
I (A)
=
V (V)
÷
(
R₁ (Ω)
+
R₂ (Ω)
)
s
Switch
example

Formulas

I
=
V
÷
(
R₁
+
R₂
)
(current
=
battery
voltage
÷
(first
bulb’s
resistance
+
second
bulb’s
resistance))
1 = 6 ÷ (2 + 4)

Assumptions

•
A circuit is a closed loop of conductor from one end of the battery to the other. Open it anywhere and the current stops.
•
Voltage (V) is the battery’s push. Current (A) is how much charge flows each second. Resistance (Ω) is how hard a part makes it to flow.
•
In series the bulbs are in one loop, so the same current goes through each and their resistances add: I = V ÷ (R₁ + R₂).
•
The switch is closed or open: tap it in the picture. Open, no current flows.

Step-by-step

WE KNOW
V = 6 V (battery voltage), R₁ = 2 Ω (first bulb’s resistance), R₂ = 4 Ω (second bulb’s resistance), s: closed
FIND
current (I)
Step 1 · Find current (I)
Use I = V ÷ (R₁ + R₂) (current = battery voltage ÷ (first bulb’s resistance + second bulb’s resistance))
Add the resistances, then divide the voltage by the total.
the switch is closed: the loop is complete
I
=
6
2 + 4
I
=
6
6
I
=
1
A
Check
Put the numbers back in. Both sides should match.
1 = 6 ÷ (2 + 4) ✓
Topic check
See what you know: 5 questions on this topic
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