The current through a 50 Ω resistor doubles from 200 mA to 400 mA. What happens to the power dissipated?
AIt doubles, from 2 W to 4 W
BIt quadruples, from 2 W to 8 W
CIt increases by 50%, from 2 W to 3 W
DIt stays the same — power depends on voltage, not current
P = I²R, so power scales as the square of current. At 200 mA: P = (0.2)² × 50 = 2 W. At 400 mA: P = (0.4)² × 50 = 8 W — four times larger. Doubling current quadruples power. This P ∝ I² relationship is the most important practical consequence of Joule heating: small increases in current cause disproportionately large increases in heat dissipation, which is why resistors have current ratings and why overloaded circuits overheat.
Question 2 Multiple Choice
Power transmission lines carry electricity over long distances. Engineers want to minimize energy lost to Joule heating during transmission. Given the same power P to transmit, which approach reduces I²R heating most effectively?
AUse thicker wires to decrease resistance R, keeping current I the same
BTransmit at high voltage and low current, so I²R losses are reduced even though P = IV stays the same
CIncrease the frequency of the alternating current
DUse lower voltage and higher current to push more power through
Since P = IV, the same power can be transmitted with high V and low I, or low V and high I. The heat loss is P_loss = I²R, which depends on the *square* of current. Halving I reduces losses by a factor of four, even if resistance is unchanged. This is why power grids use high-voltage (hundreds of kilovolts) transmission lines — the voltage is then stepped down by transformers near users. Thicker wires help but are expensive and heavy; the voltage approach is far more effective.
Question 3 True / False
The three forms P = IV, P = I²R, and P = V²/R are three different physical laws about power dissipation.
TTrue
FFalse
Answer: False
They are all the same equation expressed differently. Starting from P = IV and using V = IR (Ohm's law), substitute V = IR to get P = I(IR) = I²R, or substitute I = V/R to get P = (V/R)V = V²/R. All three forms are algebraically equivalent and describe the same physical phenomenon. The choice of form depends only on which two quantities you know directly: I and R → use I²R; V and R → use V²/R; I and V → use IV.
Question 4 True / False
A resistor with lower resistance generally dissipates less power than a resistor with higher resistance connected in the same circuit.
TTrue
FFalse
Answer: False
Whether lower R means less or more power depends on what is held fixed. If voltage V is fixed (e.g., both resistors connected to the same voltage source), P = V²/R increases as R decreases — the lower-resistance resistor dissipates *more* power. If current I is fixed (e.g., series circuit), P = I²R decreases as R decreases — the lower-resistance resistor dissipates less. Never assume the direction of the R-P relationship without specifying whether V or I is held constant.
Question 5 Short Answer
Explain microscopically what 'Joule heating' is: why does current flowing through a resistor cause the resistor's temperature to rise?
Think about your answer, then reveal below.
Model answer: In a conductor, free electrons drift under the electric field. The field accelerates each electron, giving it kinetic energy. But electrons don't travel freely — they collide with lattice ions (the fixed atomic structure). In each collision, the electron's kinetic energy is transferred to the lattice as thermal vibration (heat), and the electron starts over with low velocity. The electric field continuously does work replenishing the electrons' kinetic energy, which is continuously converted to heat through collisions. The temperature rises because the lattice vibrates more intensely. Joule heating is simply the macroscopic accounting of this microscopic cycle: field does work → electron gains kinetic energy → collision converts it to heat → repeat.
The macroscopic formula P = IV captures the rate at which the electric field does work on charges (P = dW/dt = V dq/dt = VI). The fact that this all appears as heat — not stored energy — is because resistors cannot store electrical energy the way capacitors or inductors can. All work done goes immediately and irreversibly to thermal energy.