Determine the effective current value in the circuit with C and R

In summary, the solution provided calculates the equivalent resistance (R_eq) using the formula r*Xc/(r+Xc) and obtains a value of 12.28 Ohm. However, the textbook states the correct answer as 13 A. A schematic diagram would be helpful in determining where the error lies in the solution provided. It is recommended to calculate the current in each branch separately and then add them together, taking into account that they may not be in phase with the voltage.
  • #1
Leonid92
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2
Homework Statement
The rheostat and capacitor bank are connected in parallel and connected to a sinusoidal voltage source of 220 V with a frequency of 50 Hz. The resistance of the rheostat is 20 Ohm, the capacitance of the capacitor bank is 100 μF. Determine the effective current value in the unbranched part of the circuit.
Relevant Equations
1) Xc = 1/(ω*C) = 1/(2*π*f*C)
2) I = U/resistance
Given:
U = 220 V
f = 50 Hz
r = 20 Ohm
C = 100 μF
Find: I
Solution:
1) Xc = 1/(ω*C) = 1/(2*π*f*C) = 1/(2*π*50*10^-4) = 31.83 Ohm
2) R_eq - equivalent resistance
R_eq = (r*Xc)/(r+Xc) = (20*31.83)/(20+31.83) = 12.28 Ohm
3) I = U/R_eq = 220/12.28 = 17.9 A

True answer given in the textbook is 13 A.

Could you please tell, what is wrong in the solution?
 

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  • #2
Posting a schematic diagram would help to avoid guesswork on our part. Thanks.
 
  • #3
kuruman said:
Posting a schematic diagram would help to avoid guesswork on our part. Thanks.
A schematic diagram is not given in the problem statement. I attached a sheme drawn on my own.
 
  • #4
Thank you for the schematic. I am not sure how you got R_eq. I would calculate the current in the capacitor and rheostat branches separately and then add them noting that they are not necessarily in phase with the voltage.
 
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1. What is the purpose of determining the effective current value in a circuit with C and R?

The effective current value is used to calculate the amount of current flowing through a circuit at a given time. This information is important for determining the overall performance and efficiency of the circuit.

2. How is the effective current value calculated in a circuit with C and R?

The effective current value is calculated using Ohm's Law, which states that current (I) is equal to the voltage (V) divided by the resistance (R). In a circuit with both a capacitor (C) and a resistor (R), the current value will vary over time due to the charging and discharging of the capacitor.

3. What factors can affect the effective current value in a circuit with C and R?

The effective current value can be affected by the voltage applied to the circuit, the capacitance of the capacitor, and the resistance of the resistor. Additionally, external factors such as temperature and humidity can also impact the current value.

4. How does the effective current value impact the functionality of a circuit with C and R?

The effective current value is directly related to the amount of power being used in a circuit. A higher current value means that more power is being used, which can lead to increased heat and potential circuit failure. On the other hand, a lower current value can indicate a less efficient circuit.

5. What are some practical applications of determining the effective current value in a circuit with C and R?

The effective current value is used in a variety of electronic applications, such as in power supplies, audio amplifiers, and battery charging circuits. It can also be used to monitor and troubleshoot circuits to ensure they are functioning properly and efficiently.

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