IB Physics SL B.5 Current and circuits questions

Answer: C. At A, $V/I=2.0/0.40=5.0\ \Omega$ at B, $V/I=5.0/0.80=6.25\ \Omega$ Since the resistance is not constant, the behaviour is non-ohmic. Using $P=IV$ $P_A=0.80\ \mathrm{W}$ and $P_B=4.0\ \mathrm{W}$ so $P_B/P_A=5.0$

A: obtains the power ratio but treats the unequal $V/I$ values as equal, classifying the resistor as ohmic.
B: uses the current ratio, $0.80/0.40=2.0$ as the power ratio and omits the change in potential difference.
D: assumes constant resistance and uses $P\propto I^2$ giving a ratio of $4.0$

Remember: compare $V/I$ to identify ohmic behaviour, and use $P=IV$ for power at each operating point.

Answer: B. In series, the total resistance is $3R$ so the current through $R$ is $V/(3R)$ In parallel, the $2R$ resistor has the full potential difference $V$ so its current is $V/(2R)$ The required ratio is $[V/(3R)]/[V/(2R)]=2/3$ A: uses the current through the $R$ branch in parallel, $V/R$ as the denominator. C: assumes the currents are equal because the source potential differences are equal. D: omits the $2R$ resistor when finding the series resistance, giving a series current of $V/R$ Remember: series current depends on the total resistance, while each parallel branch has the full source potential difference.