RLC Circuit Behavior with Changing Frequency, Capacitance, and Voltage

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When the source frequency increases while voltage remains constant, the ammeter reading decreases due to increased impedance from the circuit being out of resonance. Increasing capacitance also leads to a decrease in the ammeter reading, as the decrease in capacitive reactance results in higher overall impedance. Conversely, when voltage increases while the source frequency is constant, the ammeter reading increases because the circuit remains in resonance, allowing for greater current flow. The explanations provided align with the fundamental equations governing RLC circuit behavior. Overall, the analysis of the circuit's response to changes in frequency, capacitance, and voltage is accurate.
Asmaa Mohammad
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Homework Statement


In the tuned circuit below, what will happen to the reading of the ammeter when:
a) source frequency increases. The voltage is constant
b) the capacitance increases. The voltage and source frequency are constant
c) The voltage increases. Source frequency is constant.
R8vfj.jpg

Homework Equations


ƒ =1/(2π√LC) Where: C is the capacitance and L is the inductance.
I= V/Z Where: V is the voltage and Z is the impedance.
Xc=1/ωC
Xl=ωL
Z=√R²+(Xl-Xc)²

The Attempt at a Solution


Here is my solution:

a) the ammeter reading decreses.
Explanation: when source frequency varies the circuit is no longer in resonance so the impedance increases,
Z=√R²+ (Xl-Xc)²

b) the ammeter reading decreases.
Explanation: when the capacitance C increases, Xc decrease, then (Xl-Xc) increses, hence the impedance increases because:
Z= √R²+(Xl-Xc)²

c) the ammeter reading increases.
Explanation: since source frequency is constant, the circuit is still in resonance. So, increasing in voltage will increase the current becuase the impedance of the circuit will not change.

Am I correct?
 
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Asmaa Mohammad said:

Homework Statement


In the tuned circuit below, what will happen to the reading of the ammeter when:
a) source frequency increases. The voltage is constant
b) the capacitance increases. The voltage and source frequency are constant
c) The voltage increases. Source frequency is constant.
R8vfj.jpg

Homework Equations


ƒ =1/(2π√LC) Where: C is the capacitance and L is the inductance.
I= V/Z Where: V is the voltage and Z is the impedance.
Xc=1/ωC
Xl=ωL
Z=√R²+(Xl-Xc)²

The Attempt at a Solution


Here is my solution:

a) the ammeter reading decreses.
Explanation: when source frequency varies the circuit is no longer in resonance so the impedance increases,
Z=√R²+ (Xl-Xc)²

b) the ammeter reading decreases.
Explanation: when the capacitance C increases, Xc decrease, then (Xl-Xc) increses, hence the impedance increases because:
Z= √R²+(Xl-Xc)²

c) the ammeter reading increases.
Explanation: since source frequency is constant, the circuit is still in resonance. So, increasing in voltage will increase the current becuase the impedance of the circuit will not change.

Am I correct?
Yes, it is correct. Well done!
 
ehild said:
Yes, it is correct. Well done!
Thank you, ehild!
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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