Math Problem Statement

In the above RLC parallel circuit, the supply voltage is 100 volts and R is 120 Ω. If at the resonant frequency of 25 cycles the inductive reactance is 63 Ω, determine the value of the capacitive reactance and calculate (a) the supply current at resonance and (b) the currents in L and C at resonance.

Solution

In the given problem, we have an RLC parallel circuit where the supply voltage is 100 V, and the resistance R=120ΩR = 120 \, \Omega. The inductive reactance XLX_L and capacitive reactance XCX_C at resonance are both given as 63Ω63 \, \Omega. The question asks for the calculation of the supply current and the currents in the inductor and capacitor at resonance.

Given:

  • Supply Voltage Vs=100VV_s = 100 \, V
  • Resistance R=120ΩR = 120 \, \Omega
  • Inductive Reactance XL=63ΩX_L = 63 \, \Omega
  • Capacitive Reactance XC=63ΩX_C = 63 \, \Omega
  • Resonant Frequency = 25 cycles (implied resonance condition)

To Calculate:

  1. Supply Current at Resonance (IsI_s)
  2. Currents in Inductor and Capacitor (ILI_L and ICI_C)

Explanation:

  • At resonance, in a parallel RLC circuit, the inductive reactance and capacitive reactance cancel each other out because XL=XCX_L = X_C, and the net reactance of the circuit becomes zero.
  • The current through the inductor ILI_L and capacitor ICI_C will be equal in magnitude but opposite in phase. Thus, they cancel each other, and the total current supplied by the source is primarily determined by the resistive branch.

(a) Supply Current at Resonance:

At resonance, the total impedance ZtotalZ_{\text{total}} is dominated by the resistance RR, and the supply current is determined by Ohm's Law: Is=VsRI_s = \frac{V_s}{R} Is=100V120Ω=0.833AI_s = \frac{100 \, V}{120 \, \Omega} = 0.833 \, \text{A} Therefore, the supply current at resonance Is=0.833AI_s = 0.833 \, \text{A}.

(b) Currents in the Inductor and Capacitor:

The currents through the inductor and capacitor can be calculated using the reactance values: IL=VsXLI_L = \frac{V_s}{X_L} IL=100V63Ω1.59AI_L = \frac{100 \, V}{63 \, \Omega} \approx 1.59 \, \text{A} Similarly, for the capacitor: IC=VsXCI_C = \frac{V_s}{X_C} IC=100V63Ω1.59AI_C = \frac{100 \, V}{63 \, \Omega} \approx 1.59 \, \text{A} Thus, IL=1.59AI_L = 1.59 \, \text{A} and IC=1.59AI_C = 1.59 \, \text{A}.

Summary:

  • The supply current at resonance Is=0.833AI_s = 0.833 \, A.
  • The current through the inductor IL=1.59AI_L = 1.59 \, A.
  • The current through the capacitor IC=1.59AI_C = 1.59 \, A.

Let me know if you'd like more details, or if you have any questions! Here are some follow-up questions to deepen your understanding:

  1. What would happen to the supply current if the resistance was decreased?
  2. How would the resonant frequency change if the inductance or capacitance were altered?
  3. What happens to the phase relationship between voltage and current in the circuit at resonance?
  4. What are the implications of resonance in practical electrical circuits like power transmission systems?
  5. How would you calculate the impedance of the circuit if it were not at resonance?

Tip:

In parallel resonance, while the voltage across the inductor and capacitor is high, their currents cancel each other, leaving only the current through the resistor as the net supply current.

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Math Problem Analysis

Mathematical Concepts

AC Circuits
Resonance in RLC Circuits
Impedance

Formulas

I_s = V_s / R
I_L = V_s / X_L
I_C = V_s / X_C

Theorems

Resonance Condition in Parallel RLC Circuits
Ohm's Law

Suitable Grade Level

Grades 11-12