In a series LCR Circuit R = 300 Ω, L = 0.9 H, c = 2.0 µF and w = 1000 rad/sec. Then the impedance of the circuit is
In a series LCR Circuit R = 300 Ω, L = 0.9 H, c = 2.0 µF and w = 1000 rad/sec. Then the impedance of the circuit is
- 500 Ω
- 1300 Ω
- 400 Ω
- 900 Ω
The Correct Option is A
Approach Solution - 1
The impedance of a series LCR circuit can be calculated using the formula:
\( Z = \sqrt{R^2 + (X_L - X_C)^2}\)
Where \( R \) is the resistance, \( X_L \) is the inductive reactance, and \( X_C \) is the capacitive reactance.
Given values:
- R = 300 Ω
- L = 0.9 H
- C = 2.0 µF
- ω = 1000 rad/sec
The inductive reactance \( X_L \) can be calculated as follows:
\( X_L = \omega L\)
Substituting the given values:
\( X_L = (1000 \, \text{rad/sec})(0.9 \, \text{H}) = 900 \, \Omega\)
The capacitive reactance \( X_C \) can be calculated as follows:
\( X_C = \frac{1}{\omega C}\)
Substituting the given values:
\( X_C = \frac{1}{(1000 \, \text{rad/sec})(2.0 \, \mu \text{F})} = \frac{1}{2 \times 10^{-6}} \, / \, 1000 = 500 \, \Omega\)
Now, we can substitute the values of \( R \), \( X_L \), and \( X_C \) into the impedance formula:
\( Z = \sqrt{300^2 + (900 - 500)^2} = \sqrt{90000 + 160000} = \sqrt{250000} = 500 \, \Omega\)
Therefore, the impedance of the circuit is 500 Ω. So the correct option is (A) 500 Ω.
Approach Solution -2
The impedance \(Z\) of a series LCR circuit is given by:
\(Z = \sqrt{R^2 + (X_L - X_C)^2}\)
Where:
- \(R\) is the resistance
- \(X_L\) is the inductive reactance, \(X_L = \omega L\)
- \(X_C\) is the capacitive reactance, \(X_C = \frac{1}{\omega C}\)
- \(\omega\) is the angular frequency
Given:
- \(R = 300 \, \Omega\)
- \(L = 0.9 \, \text{H}\)
- \(C = 2.0 \, \mu\text{F} = 2.0 \times 10^{-6} \, \text{F}\)
- \(\omega = 1000 \, \text{rad/s}\)
Calculate the inductive reactance:
\(X_L = \omega L = (1000 \, \text{rad/s})(0.9 \, \text{H}) = 900 \, \Omega\)
Calculate the capacitive reactance:
\(X_C = \frac{1}{\omega C} = \frac{1}{(1000 \, \text{rad/s})(2.0 \times 10^{-6} \, \text{F})} = \frac{1}{0.002} = 500 \, \Omega\)
Calculate the impedance:
\(Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{(300)^2 + (900 - 500)^2}\)
\(Z = \sqrt{(300)^2 + (400)^2} = \sqrt{90000 + 160000} = \sqrt{250000} = 500 \, \Omega\)
Therefore, the impedance of the circuit is 500 Ω.
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Concepts Used:
Electrostatic Potential
The electrostatic potential is also known as the electric field potential, electric potential, or potential drop is defined as “The amount of work that is done in order to move a unit charge from a reference point to a specific point inside the field without producing an acceleration.”
SI Unit of Electrostatic Potential:
SI unit of electrostatic potential - volt
Other units - statvolt
Symbol of electrostatic potential - V or φ
Dimensional formula - ML2T3I-1
Electric Potential Formula:
The electric potential energy of the system is given by the following formula:
U = 1/(4πεº) × [q1q2/d]
Where q1 and q2 are the two charges that are separated by the distance d.