I'm studying for an electrical engineering exam, and I know how to work through this problem. (Finding the energy stored in each inductor and capacitor of the circuit in steady state.) However, I'm a little confused as to why one thing happens.
First, I understand that inductors act as short circuits in steady state and capacitors act as open circuits.
Second, I understanding using the voltage divider rule to find the current through the 6 ohm resistor. This current will also be the current flowing through the 2H inductor and the 1H inductor.
I know no current flows through the 2F Capacitor, but why does no current flow through the parallel 3 ohm resistor? In addition, because no current flows through the 3 ohm resistor at the top, I know the current through the 1 H inductor is the same as the current through the 1H inductor. Once again, why?
Thank you so much for the help.
I've attached an image of the circuit.
UPDATE:
After further thought and re-reading an explanation, I've come to establish this.
"1. Both the resistor and the capacitor are in parallel. Because the circuit is in steady state, the capacitor acts as an open circuit. No current flows through the capacitor. Therefore, no current flows through the resistor.
2. Because no current flows through the 3 ohm resistor parallel to the capacitor, the current through the 1H inductor is the same as the current through the 2H inductor.
3. Because no current flows through the 3 ohm resistor, there is no voltage drop. Therefore, there is no voltage drop across the parallel capacitor."
With this it would be found that there is no energy stored in the 2 F capacitor. Please correct me if my thought process is incorrect.
Thank you!
First, I understand that inductors act as short circuits in steady state and capacitors act as open circuits.
Second, I understanding using the voltage divider rule to find the current through the 6 ohm resistor. This current will also be the current flowing through the 2H inductor and the 1H inductor.
I know no current flows through the 2F Capacitor, but why does no current flow through the parallel 3 ohm resistor? In addition, because no current flows through the 3 ohm resistor at the top, I know the current through the 1 H inductor is the same as the current through the 1H inductor. Once again, why?
Thank you so much for the help.
I've attached an image of the circuit.
UPDATE:
After further thought and re-reading an explanation, I've come to establish this.
"1. Both the resistor and the capacitor are in parallel. Because the circuit is in steady state, the capacitor acts as an open circuit. No current flows through the capacitor. Therefore, no current flows through the resistor.
2. Because no current flows through the 3 ohm resistor parallel to the capacitor, the current through the 1H inductor is the same as the current through the 2H inductor.
3. Because no current flows through the 3 ohm resistor, there is no voltage drop. Therefore, there is no voltage drop across the parallel capacitor."
With this it would be found that there is no energy stored in the 2 F capacitor. Please correct me if my thought process is incorrect.
Thank you!
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