A pure capacitor has a capacitance of 1 farad. In a direct current (DC) circuit under steady state conditions, what is its effective resistance?

Introduction / Context:
Capacitors are widely used components in electrical and electronic circuits. Their behaviour is different in alternating current circuits and direct current circuits. This question focuses on a pure capacitor connected in a DC circuit and asks for its effective resistance in steady state. Understanding this helps clarify why capacitors block DC but pass AC signals under suitable conditions.

Given Data / Assumptions:

• We have an ideal or pure capacitor of capacitance 1 farad.
• The capacitor is connected in a DC circuit.
• We are interested in the long term or steady state behaviour, not the initial transient.
• We neglect leakage and internal resistance, treating the capacitor as ideal.

Concept / Approach:
In a DC circuit, when a capacitor is first connected to a voltage source, there is a transient current as the capacitor charges. The relationship between charge Q, capacitance C and voltage V is Q = C * V. As the capacitor charges, voltage across it increases and current decreases. After sufficient time, the capacitor becomes fully charged such that the voltage across it equals the supply voltage. At that moment, no further current flows through the circuit. An element that does not allow any steady current to pass behaves like an open circuit, which corresponds to infinite resistance in DC steady state.

Step-by-Step Solution:
Step 1: Recognise that in DC, current i in a capacitor is related to rate of change of voltage by i = C * dV / dt. Step 2: In steady state DC, the voltage across the capacitor is constant with time, so dV / dt equals zero. Step 3: Substituting into the relation gives i = C * 0 which is zero current. Step 4: A circuit element through which zero current flows under a finite applied DC voltage acts like an open circuit. Step 5: In terms of resistance, an open circuit is represented as having infinite resistance, so the effective resistance of the pure capacitor in DC is infinite.

Verification / Alternative check:
In practical DC circuits, if you place an ideal capacitor in series with a lamp and battery, the lamp may flash briefly when the capacitor starts charging, and then it goes dark permanently. The initial flash corresponds to transient current, and the final darkness corresponds to zero steady current. An ohmmeter connected across a large DC blocking capacitor also typically reads a very high resistance after charging, matching the idea that in DC steady state the capacitor path is effectively open. These observations support the conclusion of infinite effective resistance for DC.

Why Other Options Are Wrong:
Zero is incorrect because a path with zero resistance would allow maximum current, opposite to what actually happens.
1 ohm or 1 divided by 2 ohm are arbitrary finite resistances and cannot represent the complete blocking of DC current.
Equal to its reactance at 50 hertz is irrelevant, because reactance applies to AC, and in DC the frequency is zero.

Common Pitfalls:
Students sometimes confuse AC and DC behaviour. They know that capacitors can pass AC and may incorrectly assume they also allow DC current indefinitely. Others forget that once a capacitor is charged, it acts like a break in the circuit for DC. The key is to focus on steady state conditions: after transients die out, the current in a pure capacitor under DC goes to zero, which corresponds to infinite effective resistance.

Final Answer:
In a DC circuit under steady state, a pure capacitor behaves as an open circuit and has infinite effective resistance.