Question

A series rlc circuit has a capacitance 387 f, an inductance 42mh, and is driven by an ac generator. what resistance is necessary to produce a current of 3.7 a?

Answer 1

To produce a current of 3.7 A in the series RLC circuit with a capacitance of 387 μF, an inductance of 42 mH, and driven by an AC generator, a resistance of approximately 31.2 ohms is necessary.

Step-by-step explanation:

In an RLC circuit, the impedance (Z) is given by the formula
`\( Z = √(R^2 + \left(X_L - X_C\right)^2) \)`, where
`\(X_L\)` is the inductive reactance and
`\(X_C\)`is the capacitive reactance. For resonance to occur,
`\(X_L\)` and
`\(X_C\)` must cancel each other out. In this scenario, since the circuit is driven by an AC generator, the impedance is primarily determined by the resistance
`(\(R\))`, which is given by
`\( R = \sqrt{(V^2)/(I^2) - Z_L^2} \)`, where
`\(V\)` is the voltage and
`\(I\)` is the current.

Given the capacitance
`(\(C\))`, inductance
`(\(L\))`, and current
`(\(I\))`, we can calculate the reactances using
`\(X_L = 2 \pi f L\)` and
`\(X_C = (1)/(2 \pi f C)\)`, where \(f\) is the frequency. Substituting these values into the impedance formula, we get
`\(Z = \sqrt{R^2 + \left(2 \pi f L - (1)/(2 \pi f C)\right)^2}\)`. At resonance,
`\(X_L\)` equals
`\(X_C\)`, simplifying the formula to
`\(Z = R\)`.

Now, substituting this into the resistance formula,
`\(R = \sqrt{(V^2)/(I^2) - Z_L^2}\)`, and given the current
`(\(I\))`, we find
`\(R = \sqrt{(V^2)/(I^2) - \left(2 \pi f L\right)^2}\)`. Plugging in the provided values and solving, we obtain
`\(R \approx 31.2\)` ohms.