Question

The capacity of a pure capacitor is 1 farad. In dc circuits, its effective resistance will be
A. Zero
B. Infinite
C. 1 ohm
D. 0.5 ohm

Answer 1

Hint Capacitance(C) of a capacitor is the ratio of charge (Q) given and the potential (V) i.e. C=Q/V. Now, we know that the capacitive reactance is ${\chi _C} = \dfrac{1}{{2\pi \upsilon C}}$, C is the capacitance of the capacitor and ν is the frequency. As the frequency for DC current is zero then put this value in the capacitive reactance, we will get the result.

Complete step-by-step answer:
A capacitor is the combination of the two metallic plates separated by an insulating medium where the magnitude of the positive charge is spread on one plate equals to the magnitude of the negative charge on the other plate
The capacitance of capacitor is defined as the ratio of charge on the capacitor to the potential of the capacitor. $C = \dfrac{Q}{V}$, Q is the charge and V is the potential of the capacitor. It is the ability of conductor to hold a charge
As it is given that the capacity of the capacitor is 1 farad.
And we have to calculate the effective resistance of the capacitor
For this, it is given that the circuit is a DC circuit and we know that the frequency of the DC current is zero and the capacitive reactance of the capacitor is given as,
$ \Rightarrow {\chi _C} = \dfrac{1}{{2\pi \upsilon C}}$
$ \Rightarrow {\chi _C} = \dfrac{1}{{2\pi \times 0 \times C}} = \infty $
Hence, capacitors offer infinite resistance to the DC currents or we can say that, in DC circuits the effective resistance of the capacitor is infinite.

Therefore, option B is correct.

Note Capacitors can easily pass the AC current because they offer them zero resistance as they have a variable frequency but they can’t pass the DC current because they offer infinite resistance as they have zero frequency.