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In an \(L-R-C\) series circuit, the rms voltage across the resistor is 30.0 V, across the capacitor it is 90.0 V, and across the inductor it is 50.0 V. What is the rms voltage of the source?

Short Answer

Expert verified
The rms voltage of the source is 50.0 V.

Step by step solution

01

Understand the Problem

We have an L-R-C series circuit, which means that all components are connected in a single loop. We're given the rms voltages across the resistor (\(V_R = 30.0\, \text{V}\)), capacitor (\(V_C = 90.0\, \text{V}\)), and inductor (\(V_L = 50.0\, \text{V}\)). We need to find the rms voltage of the source, denoted as \(V_S\).
02

Apply the Pythagorean Theorem for Voltage

In an L-R-C series circuit, the source voltage is not simply the sum of the component voltages because they are not in phase. To find the total rms voltage of the source, we use the relationship: \[ V_S^2 = V_R^2 + (V_L - V_C)^2 \] where \(V_L - V_C\) represents the net reactance effect in the circuit.
03

Substitute the Given Values

Insert the given voltages into the equation from Step 2: \[ V_S^2 = (30.0)^2 + (50.0 - 90.0)^2 \] This becomes: \[ V_S^2 = 900 + (-40)^2 \]
04

Calculate the Expression

Compute the squared differences: \[ V_S^2 = 900 + 1600 = 2500 \]
05

Solve for the Source Voltage

Take the square root of both sides to find \(V_S\): \[ V_S = \sqrt{2500} \] \[ V_S = 50.0 \, \text{V} \]
06

Conclude the Calculation

The calculations confirm that the rms voltage of the source, \(V_S\), is 50.0 V.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

RMS Voltage
RMS, or root mean square, voltage is a way to express alternating current (AC) voltages. It represents the equivalent direct current (DC) value that would produce the same heating effect in a resistor. This is very useful since AC voltages change direction and magnitude continuously. By using RMS voltage, we can simplify calculations and comparisons. Remember that the RMS value is approximately 0.707 times the peak voltage in a sine wave. This means that if you know the peak voltage, you can easily calculate the RMS voltage using the formula:
\[ V_{rms} = rac{V_{peak}}{ ext{√2}} \]The RMS voltage lets us analyze an L-R-C series circuit more effectively by considering how each component behaves as if it were under a consistent voltage. This helps in determining the overall effect of these components on the circuit's performance.
Series Circuit
An L-R-C series circuit is a specific type of circuit where all components, such as inductors, resistors, and capacitors, are connected end-to-end in a single loop. This configuration forces the same current to flow through each component.
  • This means the total voltage across all components is the source voltage.
  • In a series arrangement, the voltages across individual components add up vectorially, not arithmetically, due to the phase differences caused by inductors and capacitors.
In analysis, you’ll often need to consider not just the magnitude, but the phase relationships between voltages across components. For instance, in our L-R-C circuit, the inductive voltage leads the current, and the capacitive voltage lags it. This phase shift results in the need to use the Pythagorean theorem in voltage calculations, as observed in the original problem.
Impedance
Impedance is a comprehensive measure of opposition that a circuit offers to the flow of alternating current. It is a combination of resistance (from resistors) and reactance (from inductors and capacitors). Impedance has both magnitude and a phase angle, which represents how the current and voltage are out of phase with each other.In a series L-R-C circuit, the presence of inductors causes the current to lag the voltage while capacitors cause the current to lead the voltage. The net reactance, represented in our problem by \( (V_L - V_C) \), affects the total impedance.
  • Reactance from inductors increases impedance while reactance from capacitors decreases it.
  • To compute overall impedance in a series circuit, it is necessary to consider both the resistive and reactive components and how they interact.
By understanding impedance, we effectively analyze how each component affects the total current flow and voltage distribution in a series circuit, allowing for the precise computation of source voltage, like in our problem where phase differences were crucial for determining the final result.

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Most popular questions from this chapter

An inductor with \(L\) = 9.50 mH is connected across an ac source that has voltage amplitude 45.0 V. (a) What is the phase angle \(\phi\) for the source voltage relative to the current? Does the source voltage lag or lead the current? (b) What value for the frequency of the source results in a current amplitude of 3.90 A?

A 0.180-H inductor is connected in series with a 90.0-\(\Omega\) resistor and an ac source. The voltage across the inductor is \(v_L\) = -(12.0 V)sin[1480 rad/s)t]. (a) Derive an expression for the voltage \(v_R\) across the resistor. (b) What is \(v_R\) at t = 2.00 ms?

You have a 200- resistor, a 0.400-H inductor, and a 6.00-F capacitor. Suppose you take the resistor and inductor and make a series circuit with a voltage source that has voltage amplitude 30.0 V and an angular frequency of 250 rad/s. (a) What is the impedance of the circuit? (b) What is the current amplitude? (c) What are the voltage amplitudes across the resistor and across the inductor? (d) What is the phase angle of the source voltage with respect to the current? Does the source voltage lag or lead the current? (e) Construct the phasor diagram

A resistor with \(R = 300\space \Omega\) and an inductor are connected in series across an ac source that has voltage amplitude 500 V. The rate at which electrical energy is dissipated in the resistor is 286 W. What is (a) the impedance Z of the circuit; (b) the amplitude of the voltage across the inductor; (c) the power factor?

An \(L-R-C\) series circuit has \(R\) = 60.0 \(\Omega\), \(L\) = 0.800 H, and \(C\) = 3.00 \(\times\) 10\(^{-4}\) F. The ac source has voltage amplitude 90.0 V and angular frequency 120 rad/s. (a) What is the maximum energy stored in the inductor? (b) When the energy stored in the inductor is a maximum, how much energy is stored in the capacitor? (c) What is the maximum energy stored in the capacitor?

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