SME
University Physics II: Electricity & Magnetism
DC circuits
Use this for “A 10 V battery with 4 Ω and a 6 V battery with 2 Ω share a 3 Ω middle branch. Find the three branch currents.”
Diagram
KCL at the junction: 2.5 + 1 = 3.5 A (in = out)
KVL round loop 1: 12 = 5 + 7 V (rises = drops)
KVL round loop 2: 9 = 2 + 7 V (rises = drops)
Each arrow shows the way its current really flows; a negative value runs against its reference.
This is an example. Type your own values to start a new problem.
I₃ (
)
=
I₁ (
)
+
I₂ (
)
ε₁ (V)
=
I₁ (
)
×
R₁ (Ω)
+
I₃ (
)
×
R₃ (Ω)
ε₂ (V)
=
I₂ (
)
×
R₂ (Ω)
+
I₃ (
)
×
R₃ (Ω)
Formulas
I₁
=
ε₁
× (
R₂
+
R₃
) −
ε₂
×
R₃
R₁
×
R₂
+
R₁
×
R₃
+
R₂
×
R₃
+2.5
=
12 × (2 + 2) − 9 × 2
2 × 2 + 2 × 2 + 2 × 2
I₂
=
ε₂
× (
R₁
+
R₃
) −
ε₁
×
R₃
R₁
×
R₂
+
R₁
×
R₃
+
R₂
×
R₃
+1
=
9 × (2 + 2) − 12 × 2
2 × 2 + 2 × 2 + 2 × 2
I₃
=
I₁
+
I₂
+3.5 = 2.5 + 1
ε₁
=
I₁
×
R₁
+
I₃
×
R₃
12 = 2.5 × 2 + 3.5 × 2
ε₂
=
I₂
×
R₂
+
I₃
×
R₃
9 = 1 × 2 + 3.5 × 2
Assumptions
•
Ideal batteries and wires; I₁ and I₂ are taken up through each battery, I₃ down through R₃.
•
Junction rule at the top node: I₃ = I₁ + I₂. Loop rule round each side: the battery’s rise equals the drops, ε₁ = I₁R₁ + I₃R₃ and ε₂ = I₂R₂ + I₃R₃.
•
Putting I₃ = I₁ + I₂ into the two loops gives two equations in I₁ and I₂, solved by Cramer’s rule; the determinant is R₁R₂ + R₁R₃ + R₂R₃.
•
A negative current flows against its arrow: the picture draws each arrow the way the current really flows.
Step-by-step
WE KNOW
ε₁ = 12 V, ε₂ = 9 V, R₁ = 2 Ω, R₂ = 2 Ω, R₃ = 2 Ω
FIND
current up through ε₁ (I₁), current up through ε₂ (I₂), current down through R₃ (I₃)
Step 1 · Find current up through ε₁ (I₁)
Use I₁ = (ε₁ × (R₂ + R₃) − ε₂ × R₃) ÷ (R₁ × R₂ + R₁ × R₃ + R₂ × R₃)
The
loops
as
a
system:
(
R₁
+
R₃
)
I₁
+
R₃
I₂
=
ε₁
and
R₃
I₁
+
(
R₂
+
R₃
)
I₂
=
ε₂
.
Cramer’s
rule
puts
ε₁
and
ε₂
in
the
I₁
column,
over
the
determinant.
I₁
=
12 × (2 + 2) − 9 × 2
2 × 2 + 2 × 2 + 2 × 2
I₁
=
12 × 4 − 9 × 2
2 × 2 + 2 × 2 + 2 × 2
I₁
=
48 − 18
4 + 4 + 4
I₁
=
30
8 + 4
I₁
=
30
12
I₁
=
+2.5
A
Step 2 · Find current up through ε₂ (I₂)
Use I₂ = (ε₂ × (R₁ + R₃) − ε₁ × R₃) ÷ (R₁ × R₂ + R₁ × R₃ + R₂ × R₃)
Cramer’s
rule
again:
ε₁
and
ε₂
go
in
the
I₂
column
this
time,
over
the
same
determinant.
I₂
=
9 × (2 + 2) − 12 × 2
2 × 2 + 2 × 2 + 2 × 2
I₂
=
9 × 4 − 12 × 2
2 × 2 + 2 × 2 + 2 × 2
I₂
=
36 − 24
4 + 4 + 4
I₂
=
12
8 + 4
I₂
=
12
12
I₂
=
+1
A
Step 3 · Find current down through R₃ (I₃)
Use I₃ = I₁ + I₂
Junction
rule
at
the
top
node:
both
battery
currents
come
down
R₃
.
I₃
=
2.5
+
1
I₃
=
+3.5
A
Check
Put the numbers back in. Both sides should match.
+2.5
=
12 × (2 + 2) − 9 × 2
2 × 2 + 2 × 2 + 2 × 2
✓
+1
=
9 × (2 + 2) − 12 × 2
2 × 2 + 2 × 2 + 2 × 2
✓
+3.5 = 2.5 + 1 ✓
12 = 2.5 × 2 + 3.5 × 2 ✓
9 = 1 × 2 + 3.5 × 2 ✓
← Previous lesson
A dielectric slid into a charged capacitor
Next lesson →
Charging a capacitor through a resistor
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
More problem types
Charging a capacitor through a resistor
Use this for “A 2 μF capacitor charges through 50 kΩ from a 12 V battery. How much charge does it hold after 0.1 s, and what current flows then?”
A battery’s internal resistance
Use this for “A 9 V battery with 0.3 Ω inside lights a 2.7 Ω bulb. Find the current, the terminal voltage and the bulb’s power.”