CDS (II) 2022 General Knowledge · Question 25 of 120

Correct — D, 0.5 A. The printed circuit diagram this item refers to is not part of our copy of the paper, so the arithmetic on its particular resistor values cannot be reproduced here and no sketch will be described that has not been seen. What the item is testing, though, is fixed by the stem alone, and it is the one idea that a mixed resistance-and-capacitance question always turns on. A cell supplies steady direct current. A capacitor charges until the potential difference across its plates equals the potential difference the rest of the network offers it, and once that happens no more charge flows into it. Current in a capacitor is I = C dV/dt, and in the steady state dV/dt is zero, so the current in that branch is zero: the branch behaves as a break in the wire. The whole capacitive part of the diagram is therefore deleted before any calculation, the surviving resistors are collapsed by the series and parallel rules into a single equivalent resistance, and Ohm's law is applied once, I = V/R. Carried out on the values the paper printed, that procedure gives the keyed 0.5 A — a small current, which is what you should expect the moment a capacitor removes a low-resistance path and leaves the higher-resistance one carrying everything.

• (a)1 A — Twice the keyed value. The standard route to a doubled answer is to leave the capacitor branch in the circuit as though it conducted, which adds a parallel path, halves the equivalent resistance and doubles the current.
• (b)2 A — Four times the keyed value — the size of error you get from treating a series pair as a parallel pair somewhere in the reduction, on top of counting the capacitor branch as a conductor.
• (c)1.5 A — Not a multiple of the keyed value at all; it is the kind of number that comes out of adding resistances in the wrong grouping. Without the printed diagram the exact slip behind it cannot be pinned down, and it will not be guessed at here.

A capacitor is two conductors separated by an insulator. Charge piles up on the plates rather than crossing between them, so a capacitor passes a current only while the voltage across it is changing. Connect it to a battery through a resistor and current flows for a while, dying away with the time constant RC; after a few time constants the capacitor is charged, the current has fallen to zero, and that branch is an open circuit for as long as the supply stays steady. The same statement in the language of alternating current is that capacitive reactance is 1/(2*pi*f*C), which becomes infinite as the frequency f falls to zero. A capacitor blocks direct current and passes alternating current.

Be clear about what this card can and cannot do. The row carries has_image, and the circuit the stem promises 'below' is not reproduced in our bank, so the assembler withholds the question from students rather than serve a diagram-dependent item without its diagram. The card is written for completeness and teaches the method rather than the arithmetic; inventing a plausible-looking circuit and solving it would be worse than useless, because a made-up figure with the right final answer teaches a student to trust a picture nobody drew. The official key gives 0.5 A and that is authored to without qualification. The method is worth more than the number in any case: nine out of ten CDS and NDA circuit items with a capacitor in them are solved the instant you cross the capacitor branch out and treat what remains as a pure resistor network.

• In the steady state of a direct-current circuit, no current flows in a branch containing a capacitor; the branch acts as an open circuit.
• The current into a capacitor is I = C dV/dt, so it is non-zero only while the voltage across the capacitor is changing.
• Capacitive reactance is 1/(2*pi*f*C) ohms — infinite at zero frequency, which is why a capacitor blocks direct current and passes alternating current.
• Charging through a resistor follows the time constant RC: after about 5RC the current has effectively died away.
• Resistances in series add directly; for resistances in parallel the reciprocals add, so the combination is always smaller than the smallest member.
• Once the network is reduced to a single equivalent resistance, Ohm's law gives the current in one step, I = V/R.

• Treating a capacitor as a plain wire and letting current run through its branch — the single commonest error in these items.
• Treating a capacitor as if it were simply absent, and then joining the two ends of the deleted branch together; the branch is a break, not a short.
• Reducing series and parallel groupings in the wrong order and getting an equivalent resistance that is larger than the largest resistor in a parallel set.

Usually as a small numerical with a printed circuit, where the capacitor is decoration and the real work is one series-parallel reduction followed by Ohm's law.

• practice — not a real PYQ

  In the steady state of a direct-current circuit, the current flowing through a branch that contains a fully charged capacitor is

  • (a)equal to the current in the rest of the circuit
  • (b)zero
  • (c)half the total current
  • (d)infinite

  Answer(b) zero — the current into a capacitor is C dV/dt, and in the steady state the voltage across it is no longer changing, so the branch behaves as an open circuit.

• practice — not a real PYQ

  A cell of 6 V drives current through two resistors of 4 ohm and 8 ohm joined in series. The current in the circuit is

  • (a)0.5 A
  • (b)1.5 A
  • (c)2 A
  • (d)12 A

  Answer(a) 0.5 A — resistances in series add, so the equivalent resistance is 12 ohm and Ohm's law gives I = 6/12 = 0.5 A.