SME
Grade 11 Science · Problem type
Resistors in parallel
Use this for “40 Ω, 60 Ω and 120 Ω are in parallel on 24 V. Find each branch current, the total current and the equivalent resistance.”
Diagram
Each
branch
gets
the
whole
battery:
I
=
V
R
.
I₁:
24
V
40
Ω
=
0.6
A
I₂:
24
V
60
Ω
=
0.4
A
I₃:
24
V
120
Ω
=
0.2
A
I = 0.6 A + 0.4 A + 0.2 A = 1.2 A
Equivalent resistance: 20 Ω
This is an example. Type your own values to start a new problem.
1
R (Ω)
=
1
R₁ (Ω)
+
1
R₂ (Ω)
+
1
R₃ (Ω)
V
Battery voltage
example
I₁
Current in R₁
calculated
I₂
Current in R₂
calculated
I₃
Current in R₃
calculated
I
Total current
calculated
Formulas
I₁
=
V
R₁
0.6
=
24
40
I₂
=
V
R₂
0.4
=
24
60
I₃
=
V
R₃
0.2
=
24
120
I
=
I₁
+
I₂
+
I₃
1.2 = 0.6 + 0.4 + 0.2
1
R
=
1
R₁
+
1
R₂
+
1
R₃
1
20
=
1
40
+
1
60
+
1
120
Assumptions
•
Each branch is connected straight across the battery, so each has the full voltage V.
•
The branch currents add at the junction: I = I₁ + I₂ + I₃.
•
The equivalent resistance is less than the smallest branch: 1/R = 1/R₁ + 1/R₂ + 1/R₃.
•
Adding a branch leaves the others’ currents the same: the new branch carries the extra current.
Step-by-step
WE KNOW
V = 24 V, R₁ = 40 Ω, R₂ = 60 Ω, R₃ = 120 Ω
FIND
current in R₁ (I₁), current in R₂ (I₂), current in R₃ (I₃), total current (I), equivalent resistance (R)
Step 1 · Find current in R₁ (I₁)
Use I₁ = V/R₁
Each branch has the whole battery voltage across it: divide by its resistance.
I₁
=
24
40
I₁
=
0.6
A
Step 2 · Find current in R₂ (I₂)
Use I₂ = V/R₂
Each branch has the whole battery voltage across it: divide by its resistance.
I₂
=
24
60
I₂
=
0.4
A
Step 3 · Find current in R₃ (I₃)
Use I₃ = V/R₃
Each branch has the whole battery voltage across it: divide by its resistance.
I₃
=
24
120
I₃
=
0.2
A
Step 4 · Find total current (I)
Use I = I₁ + I₂ + I₃
The branch currents join again: add them.
I
=
0.6
+
0.4
+
0.2
I
=
1.2
A
Step 5 · Find equivalent resistance (R)
Use 1/R = 1/R₁ + 1/R₂ + 1/R₃
Add the reciprocals of the branch resistances, then flip the sum.
R
=
1/(
1
R₁
+
1
R₂
+
1
R₃
)
R
=
1/(
1
40
+
1
60
+
1
120
)
1
R
=
0.025
+
0.01667
+
0.008333
=
0.05
R
=
1/0.05
R
=
20
Ω
Check
Put the numbers back in. Both sides should match.
0.6
=
24
40
✓
0.4
=
24
60
✓
0.2
=
24
120
✓
1.2 = 0.6 + 0.4 + 0.2 ✓
1
20
=
1
40
+
1
60
+
1
120
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
Refresh
Related lessons
Circuits: Ohm's law, series and parallel
Main lesson
Resistors in series
Use this for “A 12 V battery drives 10 Ω and 20 Ω in series. What does the ammeter read, and what is the voltage across each?”
A series-parallel circuit
Use this for “24 V drives R₁ = 4 Ω in series with 6 Ω and 12 Ω in parallel. Find the total current, the voltage across R₁ and each branch current.”
Electric power and energy
Use this for “A lamp on 120 V draws 0.5 A. What is its power and resistance, and how much energy does it use in 2 hours?”