Consider the LR circuit. We have R1=19 Ohms, R2=23 Ohms, R3=23 Ohms, V=14 Volts, L=0.09 H.
L and R2 are connected in series, Combination of L and R2 in series is connected with R3 in parallel. This whole combination is connected in series with R1 and voltage source and switch.
At t=0 the switch is closed.
(a)Immediately after the switch is closed, what are currents I1, I2, I3 ?
(b)What are I1, I2, I3 after the switch has been closed for a long time?
The switch is now opened again.
(c) What are the I1, I2, I3 currents after the switch is reopened?
a. I1=I3 = E/(R1+R3)=14/(19+23)=0.333A.
I2 = 0
b. The inductor is fully charged and
acts like a short circuit(Rc = 0).
Rt=R1 + (R2*R3)/(R2+R3)=Tot. Resistance.
Rt = 19 + (23*23)/(23+23) = 30.5 Ohms.
I1 = E/Rt = 14/30.5 = 0.46A
I2 = I3 = I1/2 = 0.46/2 = 0.23A.
c. I1 = 0
I2 = I3 = 0.23A(t = 0).
Thnx :)
C) i3 please, i3 isnt same to i2
i got it, is the negative :)
I am not getting it right with other values.
To solve this LR circuit problem, we will use the principles of circuit analysis and Ohm's Law. We will start by analyzing the circuit immediately after the switch is closed (t=0), then determine the currents after a long time has passed, and finally analyze the circuit after the switch is reopened.
(a) Immediately after the switch is closed (t=0):
Step 1: Simplify the circuit by combining the elements in series and parallel.
R2 and L are connected in series, so their equivalent resistance is R2 + L.
Step 2: Calculate the total resistance of R2 and L.
R2 = 23 Ohms
L = 0.09 H
R2 + L = 23 Ohms + 0.09 H = 23.09 Ohms (Total resistance of R2 and L)
Step 3: Combine R2 + L in series with R3.
R2 + L and R3 are connected in parallel, so their equivalent resistance is given by:
1 / (1 / (R2 + L) + 1 / R3)
1 / (1 / 23.09 + 1 / 23) = 11.26 Ohms (Equivalent resistance of R2 + L and R3)
Step 4: Combine the resulting combination in series with R1.
R1 = 19 Ohms
The total resistance in the circuit is R1 + (R2 + L || R3) = 19 + 11.26 = 30.26 Ohms.
Step 5: Calculate the total current (It) flowing in the circuit.
V = 14 Volts (Voltage source)
Using Ohm's Law (V = IR), we can determine the total current:
It = V / Rtotal = 14 / 30.26 = 0.462 A (Total current)
Step 6: Calculate the individual currents (I1, I2, and I3) using the current divider rule.
I1 = It * (R2 + L) / Rtotal = 0.462 * (23.09 / 30.26) = 0.354 A (Current through R1)
I2 = It * (R3 || R2 + L) / Rtotal = 0.462 * (11.26 / 30.26) = 0.171 A (Current through R2 + L)
I3 = It * (R3 || (R2 + L)) / Rtotal = 0.462 * (11.26 / 30.26) = 0.171 A (Current through R3)
Therefore, immediately after the switch is closed, the currents are:
I1 = 0.354 A
I2 = 0.171 A
I3 = 0.171 A
(b) After the switch has been closed for a long time:
When the switch has been closed for a long time, an LR circuit will reach its steady-state condition. In this state, the inductor behaves like a short circuit, and no current flows through it.
Since the inductor behaves like a short circuit, the circuit becomes a simple series circuit with resistors R1, R2, and R3.
Step 1: Calculate the total resistance (Rtotal) of the circuit.
Rtotal = R1 + R2 + R3 = 19 Ohms + 23 Ohms + 23 Ohms = 65 Ohms
Step 2: Calculate the total current (It) flowing in the circuit.
It = V / Rtotal = 14 / 65 = 0.215 A (Total current)
Step 3: Calculate the individual currents (I1, I2, and I3) using Ohm's Law.
I1 = It * (R1 / Rtotal) = 0.215 * (19 / 65) = 0.063 A (Current through R1)
I2 = It * (R2 / Rtotal) = 0.215 * (23 / 65) = 0.076 A (Current through R2)
I3 = It * (R3 / Rtotal) = 0.215 * (23 / 65) = 0.076 A (Current through R3)
Therefore, after the switch has been closed for a long time, the currents are:
I1 = 0.063 A
I2 = 0.076 A
I3 = 0.076 A
(c) After the switch is reopened:
When the switch is reopened, the circuit is effectively disconnected, and no current flows through any of the resistors or the inductor.
Therefore, after the switch is reopened, the currents in the circuit will be zero:
I1 = 0 A
I2 = 0 A
I3 = 0 A