A 9.2V battery is connected in series with a 33mH inductor, a 190 Ohms resistor, and an open switch.
A)What is the current in the circuit 0.140 ms after the switch is closed? in milli amps
b)How much energy is stored in the inductor at this time? in micro joules
To find the current in the circuit 0.140 ms after the switch is closed, we need to use the formula for current in an RL circuit. The formula is given as:
I(t) = (V/R) * (1 - e^(-t/(L/R)))
Where:
I(t) is the current at time t
V is the voltage provided by the battery, which is 9.2V
R is the resistance in the circuit, which is 190 Ohms
L is the inductance of the inductor, which is 33mH (0.033H)
t is the time after the switch is closed, which is 0.140 ms (0.140 * 10^-3 seconds)
Now, let's calculate the current:
I(0.140ms) = (9.2 / 190) * (1 - e^(-0.140 * 10^-3 / (0.033 / 190)))
Calculating this using a calculator:
I(0.140ms) ≈ 0.0418 A
Therefore, the current in the circuit 0.140 ms after the switch is closed is approximately 0.0418 Amperes or 41.8 milliamps.
Now, let's move on to part B to calculate the energy stored in the inductor:
The energy stored in the inductor at any given time is given by the formula:
E(t) = (1/2) * L * I^2
Where:
E(t) is the energy stored in the inductor at time t
L is the inductance of the inductor, which is 33mH (0.033H)
I is the current in the circuit at time t, which is approximately 0.0418 A
Now let's calculate the energy:
E(0.140ms) = (1/2) * (0.033) * (0.0418^2)
Calculating this using a calculator:
E(0.140ms) ≈ 2.650 μJ
Therefore, the energy stored in the inductor at 0.140 ms after the switch is closed is approximately 2.650 microjoules.
To answer these questions, we can use principles from circuit analysis and apply the basic equations involving current, voltage, and energy.
a) What is the current in the circuit 0.140 ms after the switch is closed?
To determine the current in the circuit, we need to calculate the equivalent resistance of the circuit and use Ohm's law.
First, let's find the total resistance (R_total) of the circuit. Since the circuit is in series, we can sum up the individual resistances.
R_total = 190 Ω
Next, let's determine the total inductance (L_total) of the circuit. Since the inductor is in series with other components, we can simply add up the individual inductances.
L_total = 33 mH = 33 × 10^(-3) H
Now, we can calculate the total impedance (Z_total) of the circuit using the formula:
Z_total = sqrt(R_total^2 + (2πfL_total)^2)
where f is the frequency of the circuit. However, since the switch is open, there is no frequency involved, and the impedance of the inductor is virtually infinite.
Therefore, Z_total = ∞ (infinity)
When the impedance is infinite, the current through the circuit is zero (0). Hence, 0.140 ms after the switch is closed, the current in the circuit will be zero (0).
b) How much energy is stored in the inductor at this time?
Since the current in the circuit is zero, there is no energy being stored in the inductor at this time. The energy stored in an inductor is given by the formula:
Energy = (1/2) * L * I^2
where L is the inductance and I is the current flowing through it.
As the current through the inductor is zero, the energy stored in the inductor at this time is also zero.
Therefore, the energy stored in the inductor is 0 microjoules.