Resistors R₁ = 2Ω and R₂ = 8Ω are connected in parallel. The total current entering the junction is 10A. What is the current through R₁?
A2A — because R₁/R_total = 2/10, so R₁ carries a 2/10 fraction of total current
B8A — because I₁ = I × R₂/(R₁+R₂) = 10 × 8/10
C5A — current splits equally between any two parallel branches
D2.5A — because R₁ is one-quarter of R₂, so it carries one-quarter of the current
The current divider formula for branch 1 is I₁ = I × R₂/(R₁+R₂) — note that R₂ (the OTHER branch's resistance) appears in the numerator. Here: I₁ = 10 × 8/(2+8) = 10 × 0.8 = 8A. R₁ is lower resistance, so it carries MORE current — 80% of the total. Option A is the classic mistake: using R₁ in the numerator gives the wrong result. Option C is wrong because current only splits equally when both resistances are equal.
Question 2 Multiple Choice
A student writes the current divider as I₁ = I × R₁/(R₁+R₂). What fundamental error has the student made, and what physical consequence does it produce?
AThe student used addition instead of multiplication — the formula should use R₁ × R₂
BThe student used R₁ (their own branch's resistance) in the numerator instead of R₂ — this formula would give more current to the higher-resistance branch, which contradicts Ohm's law
CThe student forgot to include the parallel equivalent resistance in the denominator
DThe formula is correct for voltage dividers but the student applied it to the wrong circuit type
The correct formula has R₂ — the OTHER branch — in the numerator. When R₁ is small (low resistance), R₂ is large, so R₂/(R₁+R₂) is close to 1, giving branch 1 most of the current. This matches intuition: lower resistance means more current. The student's formula R₁/(R₁+R₂) does the opposite — small R₁ gives a small fraction, routing most current to the higher-resistance branch. This contradicts Ohm's law and can be verified: both branches share the same voltage V, so I₁ = V/R₁, which grows as R₁ shrinks.
Question 3 True / False
If one branch of a parallel circuit has zero resistance (a short circuit), all current flows through that branch and none through the other branches.
TTrue
FFalse
Answer: True
A short circuit forces the voltage across all parallel branches to zero (V = I × 0 = 0). By Ohm's law, every other branch then carries I = 0/R = 0A regardless of its resistance. Applying the current divider formula confirms this: I₂ = I × R₁/(R₁+R₂); if R₁ = 0, then I₂ = I × 0/(0+R₂) = 0. This is why short circuits are dangerous — the short absorbs all available current.
Question 4 True / False
The current divider formula shows that each branch's current is proportional to its own resistance — a branch with twice the resistance carries twice the current.
TTrue
FFalse
Answer: False
Current is inversely proportional to resistance, not directly proportional. The formula I₁ = I × R₂/(R₁+R₂) shows that branch 1's current is determined by R₂ (the other branch) in the numerator, not R₁. If R₁ doubles, R₂/(R₁+R₂) decreases, so branch 1 carries less current. This inverse relationship follows directly from Ohm's law: both branches share the same voltage, so the higher-resistance branch must have less current (I = V/R).
Question 5 Short Answer
Explain why the current divider formula for branch 1 contains R₂ (the other branch's resistance) in the numerator, not R₁. Use the fact that parallel branches share the same voltage in your explanation.
Think about your answer, then reveal below.
Model answer: Parallel branches share the same voltage V across their terminals. By Ohm's law, I₁ = V/R₁ and I₂ = V/R₂. The total current is I = I₁ + I₂ = V/R₁ + V/R₂ = V(R₁+R₂)/(R₁R₂). Solving for V gives V = I·R₁R₂/(R₁+R₂). Substituting back: I₁ = V/R₁ = I·R₂/(R₁+R₂). The R₁ in the denominator of V/R₁ cancels with the R₁ in the numerator of V, leaving R₂ in the numerator. Physically, R₂ in the numerator means: a larger R₂ forces more current through branch 1 (since the parallel combination's voltage is higher), while a smaller R₁ also increases branch 1's current by reducing its resistance.
The derivation makes the formula feel inevitable rather than arbitrary. Once you see that both branches share voltage, the inverse-resistance relationship is immediate from Ohm's law. The 'other resistance in the numerator' is not a trick to memorize — it falls out naturally from the algebra.