
An $8\mu F$ capacitor is connected to the terminals of A.C. source whose ${V_{rms}}$ is 150 volt and the frequency is $60Hz$, the capacitive resistance is :
A) $0 \cdot 332 \times {10^3}\Omega $
B) $2 \cdot 08 \times {10^3}\Omega $
C) $4 \cdot 16 \times {10^3}\Omega $
D) $12 \cdot 5 \times {10^3}\Omega $
Answer
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Hint:In a circuit if the inductor is connected then the inductor also supplies some resistance to the flow of the current and also if the capacitor is applied then the capacitor will not only do the storing of charge but contribute to the resistance of the circuit.
Formula used: The resistance offered by the capacitor is given by,${X_C} = \dfrac{1}{{\omega C}}$ where C is the capacitance and $\omega $ is the angular frequency. The formula of the angular frequency is given by $\omega = 2\pi f$ where $\omega $ is the angular frequency and f is the frequency of the circuit.
Complete step by step answer:
It is given that the circuit has a capacitor of capacitance $8\mu F$ and the ${V_{rms}}$ is 150 volts. Also the frequency of the circuit is $60Hz$.
The angular frequency is given by,
$ \Rightarrow \omega = 2\pi f$
Where f is the frequency.
Put the value of frequency f equal to 60Hz as given in the problem.
\[ \Rightarrow \omega = 2\pi \cdot \left( {60} \right)\]
\[ \Rightarrow \omega = 120\pi \]………eq. (1)
The capacitive resistance is given by,
${X_C} = \dfrac{1}{{\omega C}}$
Where C is the capacitance and $\omega $ is the angular frequency.
$ \Rightarrow {X_C} = \dfrac{1}{{\omega C}}$………eq. (2)
Replace the value of angular frequency and capacitance from equation (1) into equation (2).$ \Rightarrow {X_C} = \dfrac{1}{{120\pi \times 0 \cdot 8 \times {{10}^{ - 3}}}}$
$ \Rightarrow {X_C} = 0 \cdot 332 \times {10^3}\Omega $
The capacitive resistance is equal to ${X_C} = 0 \cdot 332 \times {10^3}\Omega $.
The correct answer for this problem is option A.
Note: The resistance offered by the inductor and by the capacitor will depend upon the angular frequency of the a.c. source the resistance offered by the capacitor in dc source is infinite as the angular frequency is zero. When an inductor is applied in a circuit with dc source then the resistance is zero as the angular frequency is zero of the circuit.
Formula used: The resistance offered by the capacitor is given by,${X_C} = \dfrac{1}{{\omega C}}$ where C is the capacitance and $\omega $ is the angular frequency. The formula of the angular frequency is given by $\omega = 2\pi f$ where $\omega $ is the angular frequency and f is the frequency of the circuit.
Complete step by step answer:
It is given that the circuit has a capacitor of capacitance $8\mu F$ and the ${V_{rms}}$ is 150 volts. Also the frequency of the circuit is $60Hz$.
The angular frequency is given by,
$ \Rightarrow \omega = 2\pi f$
Where f is the frequency.
Put the value of frequency f equal to 60Hz as given in the problem.
\[ \Rightarrow \omega = 2\pi \cdot \left( {60} \right)\]
\[ \Rightarrow \omega = 120\pi \]………eq. (1)
The capacitive resistance is given by,
${X_C} = \dfrac{1}{{\omega C}}$
Where C is the capacitance and $\omega $ is the angular frequency.
$ \Rightarrow {X_C} = \dfrac{1}{{\omega C}}$………eq. (2)
Replace the value of angular frequency and capacitance from equation (1) into equation (2).$ \Rightarrow {X_C} = \dfrac{1}{{120\pi \times 0 \cdot 8 \times {{10}^{ - 3}}}}$
$ \Rightarrow {X_C} = 0 \cdot 332 \times {10^3}\Omega $
The capacitive resistance is equal to ${X_C} = 0 \cdot 332 \times {10^3}\Omega $.
The correct answer for this problem is option A.
Note: The resistance offered by the inductor and by the capacitor will depend upon the angular frequency of the a.c. source the resistance offered by the capacitor in dc source is infinite as the angular frequency is zero. When an inductor is applied in a circuit with dc source then the resistance is zero as the angular frequency is zero of the circuit.
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