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Question: A condenser of capacity \(1pF\) is connected to an A.C source of \(220V\) and \(50H\) frequency. The...

A condenser of capacity 1pF1pF is connected to an A.C source of 220V220V and 50H50H frequency. The current flowing in the circuit will be
(A) 6.9×108A6.9 \times {10^{ - 8}}A
(B) 6.9A6.9A
(C) 6.9×106A6.9 \times {10^{ - 6}}A
(D) Zero

Explanation

Solution

It is a direct question. We must know the basic concept of electromagnetic Induction and alternating current. Use the equation derived in the concept of effective current in an a.c circuit with capacitor.

Complete step by step answer:
When an circuit is an a.c circuit with an capacitor, the Irms or Ieff{{\text{I}}_{{\text{rms}}}}{\text{ or }}{{\text{I}}_{{\text{eff}}}} value of the effective current in the circuit is given by
Irms=ErmsXc 1\Rightarrow {I_{rms}} = \dfrac{{{E_{rms}}}}{{{X_c}}}{\text{ }} \to {\text{1}}
Where,
Irms{I_{rms}} is the effective current in the circuit.
Erms{E_{rms}} is the effective emf (voltage) of the circuit .
XC{X_C} is the capacitive reactance.
The capacitive resistance is the resistance offered by the capacitor .Its unit is ohm. The capacitive resistance is given by
XC=1ωc=12πνc\Rightarrow {X_C} = \dfrac{1}{{\omega c}} = \dfrac{1}{{2\pi \nu c}}
ω\omega is the angular frequency whose value is 2πν2\pi \nu [ ν\nu is the frequency ]
C is the capacitance
The equation 1 becomes,
Irms=Erms12πνC 2\Rightarrow {I_{rms}} = \dfrac{{{E_{rms}}}}{{\dfrac{1}{{2\pi \nu C}}}}{\text{ }} \to 2
Given that,
The capacitance of the capacitor, C=1pF=1×1012FC = 1pF = 1 \times {10^{ - 12}}F
The a.c offers an effective voltage, Erms=220V{E_{rms}} = 220V
The frequency of the circuit, ω=50H\omega = 50H
We have to find the current in the circuit.
From the equation 2
The value of the effective current in the circuit is given by
Irms=Erms12πνC 2\Rightarrow {I_{rms}} = \dfrac{{{E_{rms}}}}{{\dfrac{1}{{2\pi \nu C}}}}{\text{ }} \to 2
Substitute the known values
Irms=22012×3.14×50×1×1012\Rightarrow {I_{rms}} = \dfrac{{220}}{{\dfrac{1}{{2 \times 3.14 \times 50 \times 1 \times {{10}^{ - 12}}}}}}
Irms=220×2×3.14×50×1×1012\Rightarrow {I_{rms}} = 220 \times 2 \times 3.14 \times 50 \times 1 \times {10^{ - 12}}
Irms=6.9×108A\Rightarrow {I_{rms}} = 6.9 \times {10^{ - 8}}A
The current flowing in the circuit is, Irms=6.9×108A{I_{rms}} = 6.9 \times {10^{ - 8}}A

Hence the correct answer is option A) 6.9×108A6.9 \times {10^{ - 8}}A

Additional information:
The rms value of alternating current is defined as that value of the steady current, which when passed through a resistor for a given time, will generate the same amount of heat as generated by an alternating current when passed through the same resistor for the same time. The rms value is also called effective value. It is denoted by Irms or Ieff{{\text{I}}_{{\text{rms}}}}{\text{ or }}{{\text{I}}_{{\text{eff}}}}
Irms=I02\Rightarrow {I_{rms}} = \dfrac{{{I_0}}}{{\sqrt 2 }}
Where, I0{I_0} is the maximum value of current.
The same concept applies for emf.
The rms value of alternating emf is given by
Erms=E02\Rightarrow {E_{rms}} = \dfrac{{{E_0}}}{{\sqrt 2 }}
Where, E0{E_0} is the maximum value of induced emf

Note: Alternating current (a.c) varies continuously with time and its average value over one complete cycle is zero since the crust and tuft gets cancelled with each other. Hence it is measured by its rms value. rms value is the short form of root mean square value.