Questions: Ohm's Law and Resistance

Question 1

A 10Ω resistor has 5V across it. The voltage is then doubled to 10V. What happens to the current through the resistor?

• A. It stays the same — current depends only on the power supply's internal resistance
• B. It doubles, from 0.5A to 1.0A
• C. It quadruples, since power increases as the square of voltage
• D. It halves — higher voltage increases effective resistance

Answer: (B)

By Ohm's Law, I = V/R. With R constant at 10Ω: at 5V, I = 0.5A; at 10V, I = 1.0A. The linear proportionality between V and I is the defining property of an ohmic resistor — double the voltage, double the current. Option D is the classic misconception: for an ohmic material, resistance is a fixed material property and does not change with applied voltage. Non-ohmic behavior (where effective resistance changes with voltage) would be a different kind of device entirely.

Question 2

Two resistors — 100Ω and 10Ω — each have 10V applied across them. Which dissipates more power, and by what factor?

• A. The 100Ω resistor, because greater resistance means more power dissipated
• B. The 10Ω resistor, by a factor of 10
• C. They dissipate equal power, since the voltage across each is identical
• D. The 10Ω resistor, by a factor of 100

Answer: (B)

Use P = V²/R when voltage is known. P₁₀₀ = (10)²/100 = 1W; P₁₀ = (10)²/10 = 10W. The 10Ω resistor dissipates 10× more power. This is counterintuitive: lower resistance means more power at constant voltage, because less resistance allows larger current to flow. Option A confuses P = I²R (where more resistance means more power at constant current) with P = V²/R (where more resistance means less power at constant voltage). The formula to use depends on whether current or voltage is held fixed by the circuit.

Question 3

A longer wire of the same material and cross-sectional area has lower resistance than a shorter wire of the same material.

• T. True
• F. False

Answer: False

Resistance is proportional to length: R = ρL/A. A longer wire has higher resistance — more resistive material for charge carriers to traverse. Resistance is inversely proportional to cross-sectional area — a thicker wire has lower resistance than a thinner one of the same length and material. This is why undersized wiring for high-current loads causes voltage drop and overheating: the wire's resistance is too high for the current being pushed through it.

Question 4

Conductance G = 1/R is useful in parallel circuit analysis because conductances in parallel add directly, just as resistances add directly in series.

• T. True
• F. False

Answer: True

This duality is exact. In series circuits, the same current flows through all elements, so resistances add: R_total = R₁ + R₂ + …. In parallel circuits, the same voltage appears across all branches, so conductances add: G_total = G₁ + G₂ + …. Both follow from V = IR — it's a matter of which quantity is shared. Choosing to work in resistance or conductance is a computational convenience depending on which topology dominates the circuit being analyzed.

Question 5

Explain why you should use P = I²R in some situations and P = V²/R in others. What determines which form is appropriate?

Model answer: Both forms are derived from P = VI combined with V = IR. The choice depends on which quantity is fixed by the circuit configuration. In a series circuit, the same current flows through all elements, so I is known — use P = I²R. In a parallel circuit, the same voltage appears across all branches, so V is known — use P = V²/R. Using the wrong formula implicitly assumes a different circuit topology and gives an incorrect result.

A concrete example: a 100Ω and 10Ω resistor in series carrying 1A — use P = I²R: the 100Ω dissipates 100W, the 10Ω dissipates 10W. The same resistors in parallel with 10V applied — use P = V²/R: the 100Ω dissipates 1W, the 10Ω dissipates 10W. Same resistors, completely different power distribution, because the circuit topology changes which quantity is fixed.