A 12 V battery drives two resistors in parallel: 4.0 Ω and 6.0 Ω. One routine gives every current in the circuit: both branch currents and the supply current.
The routine leans on the full-voltage rule. Check it first:
Two fans run in parallel from a 24 V supply. What is the voltage across each fan?
Step 1: write down what each branch has across it. Every branch has the full source voltage: 12 V.
Step 2: find each branch current with I = V/R, one branch at a time.
For the 4.0 Ω branch: I = 12/4.0 = 3.0 A.
For the 6.0 Ω branch: I = 12/6.0 = 2.0 A.
Step 3: add the branch currents to get the supply current: 3.0 A + 2.0 A = 5.0 A.
Run the sense checks: the smaller resistance (4.0 Ω) carries the bigger current, and the supply current is bigger than either branch current.
An 18 V supply drives a 6.0 Ω resistor and a 9.0 Ω resistor in parallel. Find both branch currents and the supply current.
Write down the values in the question:
V = 18 V (across every branch)
R₁ = 6.0 Ω
R₂ = 9.0 Ω
I₁ = ?, I₂ = ?, supply current = ?
Write down the equation for each branch:
I = V/R
Substitute in the values for the 6.0 Ω branch:
I₁ = 18/6.0
Simplify:
I₁ = 3.0 A
Substitute in the values for the 9.0 Ω branch:
I₂ = 18/9.0
Simplify:
I₂ = 2.0 A
Add the branch currents:
supply current = 3.0 A + 2.0 A = 5.0 A
I₁ = 3.0 A, I₂ = 2.0 A, and the supply delivers 5.0 A.
Now the same routine on a fresh circuit, one step per question: a 20 V supply drives 5.0 Ω and 10 Ω resistors in parallel.
A 20 V supply drives 5.0 Ω and 10 Ω resistors in parallel. Current in the 5.0 Ω branch, in A?
The same circuit: 20 V across 5.0 Ω and 10 Ω in parallel. Current in the 10 Ω branch, in A?
Still the 20 V supply with 5.0 Ω and 10 Ω in parallel. What current does the supply deliver, in A?
The 9.0 Ω resistor is swapped for a 3.0 Ω resistor. Find the new branch current and the new supply current.
Write down the values in the question:
V = 18 V (across every branch)
R₂ = 3.0 Ω
I₂ = ?, supply current = ?
Write down the equation:
I = V/R
Substitute in the values for the 3.0 Ω branch:
I₂ = 18/3.0
Simplify:
I₂ = 6.0 A
Add the branch currents (the 6.0 Ω branch still carries 3.0 A):
supply current = 3.0 A + 6.0 A = 9.0 A
I₂ = 6.0 A, and the supply now delivers 9.0 A.
A 15 V supply drives a 3.0 Ω resistor and a 5.0 Ω resistor in parallel. Calculate the current the supply delivers, in A.
A 16 V supply drives an 8.0 Ω resistor and a 4.0 Ω resistor in parallel. Which pair gives the currents in the 8.0 Ω branch and the 4.0 Ω branch, in that order?
For a 12 V supply across parallel resistors of 2.0 Ω and 4.0 Ω, a student adds the resistances and finds I = 12/6.0 = 2.0 A for each branch. What is the first wrong step?