Questions: Current Divider Principle and Applications

Question 1

Two resistors R₁ = 10 Ω and R₂ = 90 Ω are connected in parallel with total current I_total = 100 mA entering the node. What is the current through R₂?

• A. 90 mA — R₂ is larger so it carries the proportionally larger share of current
• B. 10 mA — the current divider formula gives I₂ = I_total × R₁/(R₁+R₂) = 100 × 10/100
• C. 50 mA — current splits equally between parallel branches regardless of resistance
• D. 9 mA — the formula is I₂ = I_total × R₂/(R₁+R₂) = 100 × 90/100

Answer: (B)

The current divider formula is I₂ = I_total × R₁/(R₁+R₂). Note that R₁ (the *other* resistor) appears in the numerator for I₂. Here: I₂ = 100 mA × 10/(10+90) = 100 × 0.1 = 10 mA. The remaining 90 mA flows through R₁. Current divides inversely with resistance: R₂ is 9× larger than R₁, so it carries 1/9 the current. Option D uses the wrong formula — placing the branch's own resistance in the numerator is the voltage divider pattern, not the current divider.

Question 2

One branch of a parallel circuit has its resistance increased by a factor of 10. What happens to the current through the other branch?

• A. The current through the other branch decreases because total current must be shared between fewer effective branches
• B. The current through the other branch increases, since the high-resistance branch now diverts less current away
• C. The current through the other branch is unchanged — parallel branches are completely independent
• D. Both branches carry less current because increasing one resistance raises the total equivalent resistance

Answer: (B)

In a current-driven parallel circuit, total current is fixed. If one branch's resistance increases, that branch carries much less current. By KCL, all current entering the node must leave through some branch — so the remaining branches absorb more. The easier-path branch gets a larger fraction of total current when its competitor becomes harder. Option C is wrong: the branches share the same voltage, but the distribution of current between them shifts when resistance changes. Option D confuses a current source (fixed total current) with a voltage source scenario.

Question 3

In a parallel circuit, the branch with the highest resistance carries the most current.

• T. True
• F. False

Answer: False

Current divides *inversely* with resistance. Both parallel branches see the same terminal voltage V. By Ohm's law, I = V/R, so the branch with lower resistance carries more current. Higher resistance is a harder path for current to flow through, so less current takes that route. This is the defining feature of the current divider — it is the opposite of what intuition from series circuits might suggest, where higher resistance means more voltage drop.

Question 4

The current divider formula I₂ = I_total × R₁/(R₁+R₂) uses R₁ in the numerator for branch 2's current because both branches share the same voltage, and R₁ determines how large that shared voltage is.

• T. True
• F. False

Answer: True

The appearance of R₁ in the numerator for I₂ is not arbitrary — it reflects the underlying physics. The shared voltage V = I_total × R_eq = I_total × R₁R₂/(R₁+R₂). Since I₂ = V/R₂, substituting gives I₂ = I_total × R₁/(R₁+R₂). A larger R₁ means higher equivalent resistance and higher shared voltage, which drives more current through R₂. The 'other' resistance controls the voltage that in turn drives current through the branch of interest.

Question 5

Explain in physical terms why the current divider formula places the opposite resistor's value in the numerator (I₂ involves R₁, not R₂).

Model answer: Parallel branches share the same voltage. Current through any branch is I = V/R. Since both branches see identical voltage, the branch with lower resistance draws more current — it's the easier path. The formula I₂ = I_total × R₁/(R₁+R₂) naturally encodes this: if R₂ is small relative to R₁, the ratio R₁/(R₁+R₂) approaches 1, and branch 2 gets nearly all the current. If R₂ is large, the ratio is small, and branch 2 gets little current. R₁ in the numerator reflects that a large competing resistance means more voltage is available to drive current through branch 2.

The inversion makes sense from the limiting cases. If R₂ → 0 (short circuit), branch 2 should get all the current: R₁/(R₁+0) = 1, so I₂ = I_total. If R₂ → ∞ (open circuit), branch 2 should get no current: R₁/(R₁+∞) → 0, so I₂ = 0. The formula's behavior at extremes confirms the physical intuition — current takes the easiest available path.