Physics

1. 1.

   Resistors 6.0 ohm and 3.0 ohm are connected in parallel across an ideal 12 V source. Calculate equivalent resistance and total current, to 2 significant figures.

   [3 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Each branch has the full 12 V, giving branch currents 2.0 A and 4.0 A.
   2. Their sum agrees with dividing the source voltage by equivalent resistance.

   Marking points

   • 1/R = 1/6.0 + 1/3.0.
   • R = 2.0 ohm.
   • Total current is 6.0 A.

   Examiner tip: A parallel equivalent must be smaller than the smallest branch resistance.

2. 2.

   An ideal 200 microfarad capacitor is charged to 12.0 V. Calculate its charge and stored energy, to 3 significant figures.

   [3 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. The micro prefix introduces a factor 10^-6 before using SI equations.
   2. Charging voltage rises from zero, so stored energy is half QV, not QV.

   Marking points

   • C = 2.00e-4 F and Q = CV.
   • Q = 0.00240 C.
   • Energy = CV^2/2 = 0.0144 J.

   Examiner tip: Square voltage in the energy formula, but not in Q = CV.

3. 3.

   A cell gives terminal voltage 5.40 V at current 0.300 A and 4.80 V at 0.600 A. Find its emf and internal resistance, assuming both constant.

   [3 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Subtracting the equations eliminates emf and isolates the slope magnitude of the V-I line.
   2. Add the internal drop back to either terminal voltage; both measurements recover 6.00 V.

   Marking points

   • Use V = E - Ir for both measurements.
   • r = (5.40 - 4.80)/(0.600 - 0.300) = 2.00 ohm.
   • E = 5.40 + 0.300*2.00 = 6.00 V.

   Examiner tip: Internal resistance is minus the V-versus-I gradient, not V/I at one point.

4. 4.

   A 100 microfarad capacitor initially at 10.0 V discharges through 20.0 kilohm. Find voltage and current magnitude after 3.00 s. Assume an ideal capacitor and constant resistance; give 3 significant figures.

   [3 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. The voltage decays exponentially; use the original voltage as the prefactor.
   2. Use unrounded voltage to calculate current, then convert amperes to milliamperes.

   Marking points

   • Time constant RC = 2.00 s.
   • V = 10.0e^(-3.00/2.00) = 2.23 V.
   • Current magnitude V/R = 0.112 mA.

   Examiner tip: The exponent t/RC is dimensionless; convert kiloohms and microfarads first.

5. 5.

   A 12.0 V ideal supply feeds a divider: 2.00 kilohm above the output node and 4.00 kilohm below it to ground. A 4.00 kilohm load is added from output to ground. Find loaded output voltage and supply power. Compare with the unloaded voltage.

   [3 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. The load is parallel to the lower divider resistor, reducing the lower effective resistance.
   2. The loaded series total is 4.00 kilohm; the equal halves split the voltage equally. Without a load the lower fraction was 4/6.

   Marking points

   • The lower equivalent is 4.00 || 4.00 = 2.00 kilohm.
   • Loaded output is 6.00 V; source current is 3.00 mA.
   • Supply power is 0.0360 W; unloaded output was 8.00 V.

   Examiner tip: The load changes the divider ratio; do not reuse the unloaded output.

6. 6.

   An isolated 300 microfarad capacitor at 12.0 V is connected in parallel, like terminals together, to an uncharged 600 microfarad capacitor. Find final voltage and decrease in stored energy. Ideal capacitors; negligible leakage; allow transient dissipation in wires.

   [3 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. The isolated pair shares its original charge, not its original voltage, until both have equal potential difference.
   2. The energy decrease is dissipated during the transient; electrostatic energy conservation is not an appropriate constraint here.

   Marking points

   • Total charge conserved: 300e-6*12.0 = (900e-6)V.
   • Final voltage is 4.00 V.
   • Initial energy 0.0216 J; final energy 0.00720 J; decrease 0.0144 J.

   Examiner tip: Conserve charge for redistribution; do not force capacitor energy to stay constant.

Original, unofficial practice questions from SubjectScout — not official past-paper questions.