[SOLVED] Current and Voltage of an Inductor circuit

[preethi19]

Hi i have attached two images one with capacitor and one with inductors and resistors.


For the capacitor and resistors circuit since the supply is DC the current through the capacitor is 0 but i was able to determine the charge Q and the voltage across both the capacitors with the formula Q=CV. Similarly for an inductor circuit i would like to calculate the current through the inductor and voltage across the inductor. I only know the formula V=L dI/dt. But the current in the image is DC and not varying so is der any simple formula that i can calculate I and V in the DC circuit of the inductor. Kindly help!!! Thank you!!!

 

[albbg]

You just have to replace each inductor with a short circuit, than calculate the current in each branch. The current on each inductor will be equal to that of the corresponding branch.
Of course the result is valid at steady state (after the initial transient).

 

[preethi19]

Thank you for your reply!!! Calculating current through the inductors as you told is the current in each branch. How can i determine the voltage across the inductor??? Pls help!!!

 

[chuckey]

The maximum current is 12/50 A because all the inductors have zero resistance. This current flows through L1. I cannot say how the current splits through the inductors because they all have zero ohms, likewise the voltage drop across all the inductors is 0V.
Frank

 

[preethi19]

I simulated for the circuit and found the following values from the simulator. The voltage across each inductor is 0. The current through
L1 = 240mA
L2 = 161.31mA
L3 = 78.69mA
L4 = 78.69mA
L5 = 31.48mA
L6 = 47.21mA

The current through resistor R1 is 240mA and i can see the current through inductor L1 must be the same 240mA. But how are the current values calculated for the rest of the inductors???

Also i found this explanation "In case of inductor, vl(t)=L *dil(t)/dt, vl(t) is the voltage across the inductor, hence when circuit is closed there is huge di/dt in transisent state and inductor will act as huge resistor. But as the current becomes constant at steady state, di/dt = 0, V(l) = 0 which means voltage across inductor is zero hence short."
In the simulated circuit i can see the voltage across all inductors are 0 at steady state and by the explanation i found then the inductor becomes short circuit. But i see current flow through the circuit. How is this possible. If short circuit there should be no current flow too right. Bit confusing. Pls help!!!

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So if not using ohms is the way to calculate then wat is the other way to determine this???

 

[asdf44]

No that's not correct. A wire is a short circuit and wires carry current!

If it helps, in real life nothing is a short circuit, it's all a small R. Consider all the inductors to be 1 mili-ohm and you'd see they have voltage and therefor current as well.


If you're new at this the symbols for C's and L's tell you the DC state. Caps, which are drawn with a gap are literally 'open' to DC current and provide no path. Inductors, drawn as a wire, are shorts just like wires.

EDIT: The simulator may be adding in series resistance without telling you. In general, there is no answer for the circuit you posted. It's either a trick question or a mistake.

 

[albbg]

preethi19;1524097 But i see current flow through the circuit. How is this possible. If short circuit there should be no current flow too right. Bit confusing. Pls help!!!

You are confusing short-circuit with open circuit.

Short-circuit between to nodes A and B, means that A and B are at the same potential, then the voltage (that is the potential difference) V(A,B) = 0 while the current can flows from A to B
Open-circuit between to nodes A and B, means that A and B are disconnected one each other, then in general V(A,B) <> 0 while the current cannot flows

Suppose you have a resistor R connected between A and B. We can apply the Ohm's law, so V(A,B) = R*I(R) or I(R) = V(A,B)/R
Short-circuit means R = 0 and as you can see V(A,B) = 0*I(R) = 0
Open-circuit means R-->inf, thus I(R) = lim(R-->inf) { V(A,B)/R} = 0