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Question: A \(10\mu F\) capacitor is connected across a \(200V\), \(50Hz\)A.C. supply. The peak current throug...

A 10μF10\mu F capacitor is connected across a 200V200V, 50Hz50HzA.C. supply. The peak current through the circuit is:
0.6A0.6A
0.62A0.6\sqrt{2}A
0.062A0.06\sqrt{2}A
0.6πA0.6\pi A

Explanation

Solution

The energy source of this circuit is AC (alternating current). Also, it is a purely capacitive circuit. This means that only a capacitor is present in the circuit along with the alternating current source and no resistor or inductor is present. Therefore, only the capacitor is consuming energy and the current faces resistance only by the capacitive reactance of the capacitor.

Complete step-by-step solution:
Impedance is the total resistance present against the current in the circuit.
Impedance of a LCR circuit is given as:
Z=R2+(XL−XC)2Z=\sqrt{{{R}^{2}}+{{\left( {{X}_{L}}-{{X}_{C}} \right)}^{2}}}
Where,
R=R= Resistance of the resistor present
XL={{X}_{L}}=Inductive reactance of the inductor present. It is further given as XL=ωL{{X}_{L}}=\omega L
ω=2πν\omega =2\pi \nu
⇒XL=2πνL\Rightarrow {{X}_{L}}=2\pi \nu L
Where, ν=\nu = frequency of the alternating current provided to the circuit and L=L=inductance of inductor
XC={{X}_{C}}=Capacitive reactance of the capacitor present. It is further given as XC=1ωC{{X}_{C}}=\dfrac{1}{\omega C}
ω=2πν\omega =2\pi \nu , ⇒XC=12πνC\Rightarrow {{X}_{C}}=\dfrac{1}{2\pi \nu C} ………….. equation (1)
Where, ν=\nu =frequency of the alternating current provided to the circuit and C=C=capacitance of capacitor


Since, no resistor or inductor is present in the circuit,
Therefore,
Z=0+(0−XC)2 ⇒Z=XC \begin{aligned} & Z=\sqrt{0+{{\left( 0-{{X}_{C}} \right)}^{2}}} \\\ & \Rightarrow Z={{X}_{C}} \\\ \end{aligned}
Now,
The voltage,VVof the a.c. circuit is given as:
V=IZV=IZ
Here, I=I=current in circuit and Z=XCZ={{X}_{C}}
⇒XC=VI\Rightarrow {{X}_{C}}=\dfrac{V}{I}
From equation (1),
⇒12πνC=VI\Rightarrow \dfrac{1}{2\pi \nu C}=\dfrac{V}{I}
Given that voltage = 200 volts, frequency = 50Hz50Hz and capacitance =10μF=10×10−6F=10−5F10\mu F=10\times {{10}^{-6}}F={{10}^{-5}}F
⇒12π(50)(10−5)=200I ⇒I=π200(100)(10−5) ⇒I=2π(10−1) \begin{aligned} & \Rightarrow \dfrac{1}{2\pi \left( 50 \right)\left( {{10}^{-5}} \right)}=\dfrac{200}{I} \\\ & \Rightarrow I=\pi 200\left( 100 \right)\left( {{10}^{-5}} \right) \\\ & \Rightarrow I=2\pi \left( {{10}^{-1}} \right) \\\ \end{aligned}
∴I=0.2πA\therefore I=0.2\pi A ……………. Equation (2)
The voltage given and current that we have just calculated are the rms voltage and rms current respectively (where rms stands for root mean square).
The rms voltage is related to the peak voltage as: Vrms=Vpeak2{{V}_{rms}}=\dfrac{{{V}_{peak}}}{\sqrt{2}}
Similarly, the rms current is related to the peak current as: Irms=Ipeak2{{I}_{rms}}=\dfrac{{{I}_{peak}}}{\sqrt{2}}
⇒Ipeak=2Irms\Rightarrow {{I}_{peak}}=\sqrt{2}{{I}_{rms}}
From equation (2), we get Irms=0.2πA{{I}_{rms}}=0.2\pi A
⇒Ipeak=2(0.2π)\Rightarrow {{I}_{peak}}=\sqrt{2}\left( 0.2\pi \right)
Substituting π=3.14,\pi =3.14,
⇒Ipeak=2(0.2)(3.14) ⇒Ipeak=2(0.628) ∴Ipeak≈0.62A \begin{aligned} & \Rightarrow {{I}_{peak}}=\sqrt{2}\left( 0.2 \right)\left( 3.14 \right) \\\ & \Rightarrow {{I}_{peak}}=\sqrt{2}\left( 0.628 \right) \\\ & \therefore {{I}_{peak}}\approx 0.6\sqrt{2}A \\\ \end{aligned}
Hence, the peak current is 0.62A0.6\sqrt{2}A.

Therefore, the correct option is (B)0.62A0.6\sqrt{2}A.

Note:
The voltage rating coming to our households is the rms voltage of the alternating current. Since the alternating current occurs as sine or cosine function of time, the value of this current is not constant and keeps fluctuating. This gives rise to the need of a more stable measurement such as rms current.