SME
Circuit Analysis I · Problem type
Two resistors in parallel
Use this for “A 6 Ω and a 3 Ω resistor are in parallel across 12 V. Find each current.”
Diagram
KCL at the junction: 6 = 2 + 4 A (in = out)
This is an example. Type your own values to start a new problem.
Req (Ω)
=
R₁ (Ω)
×
R₂ (Ω)
R₁ (Ω)
+
R₂ (Ω)
I₁ (A)
=
V (V)
R₁ (Ω)
I₂ (A)
=
V (V)
R₂ (Ω)
I
Total current
calculated
Formulas
Req
=
R₁
×
R₂
R₁
+
R₂
2
=
6
×
3
6 + 3
I₁
=
V
R₁
2
=
12
6
I₂
=
V
R₂
4
=
12
3
I
=
I₁
+
I₂
6 = 2 + 4
I₁
=
I
×
R₂
R₁
+
R₂
2
=
6
×
3
6 + 3
V
=
I
×
Req
12 = 6 × 2
Assumptions
•
Both resistors are connected across the source, so each sees the full source voltage V.
•
Kirchhoff’s current law: the current from the source splits between the branches, I = I₁ + I₂.
•
The equivalent resistance Req is less than the smaller resistor; wires have no resistance.
Step-by-step
WE KNOW
V = 12 V, R₁ = 6 Ω, R₂ = 3 Ω
FIND
equivalent resistance (Req), current in R₁ (I₁), current in R₂ (I₂), total current (I)
Step 1 · Find equivalent resistance (Req)
Use Req = R₁ × R₂ ÷ (R₁ + R₂)
Product
over
sum:
1/Req
=
1
R₁
+
1
R₂
for
two
resistors
in
parallel.
Req
=
6
×
3
6 + 3
Req
=
6
×
3
9
Req
=
18
9
Req = 2 Ω
Step 2 · Find current in R₁ (I₁)
Use I₁ = V ÷ R₁
Ohm’s
law
for
R₁
,
with
the
full
source
voltage
across
it.
I₁
=
12
6
I₁
=
2
A
Step 3 · Find current in R₂ (I₂)
Use I₂ = V ÷ R₂
Ohm’s
law
for
R₂
,
with
the
full
source
voltage
across
it.
I₂
=
12
3
I₂
=
4
A
Step 4 · Find total current (I)
Use I = I₁ + I₂
Kirchhoff’s current law at the top junction: the branch currents add.
I
=
2
+
4
I
=
6
A
Check
Put the numbers back in. Both sides should match.
2
=
6
×
3
6 + 3
✓
2
=
12
6
✓
4
=
12
3
✓
6 = 2 + 4 ✓
2
=
6
×
3
6 + 3
✓
12 = 6 × 2 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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Ohm's and Kirchhoff's laws
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