A 1.0 KΩ resistor is connected in series with a 10.0 mH inductor, a 30.0 V battery, and an open switch. At time t=0, the switch is suddenly closed. What are the voltage drops VR and VL 20.00 µs after the switch is closed?
VL = Vap / e^t/T,
T=L/R=10*10^-3h/1*10^-3Ohms=10*10^-6s.
t/T = 20*10^-6s / 10*10^-6s = 2.
VL = 30 / e^2 4.06 volts.
VR = 30 - 4.06 = 25.94 volts.
To determine the voltage drops VR and VL 20.0 µs after the switch is closed, we can use the formulas for voltage drop across a resistor and an inductor in a series circuit.
1. Calculate the current (I) through the circuit using Ohm's Law, which states that V = IR, where V is the voltage and R is the resistance:
V = 30.0 V (given)
R = 1.0 KΩ = 1000 Ω (as 1 KΩ = 1000 Ω)
I = V / R = 30.0 V / 1000 Ω = 0.030 A
2. Calculate the time constant (τ) of the circuit, which is the product of the resistance and inductance:
L = 10.0 mH = 10.0 x 10^-3 H (as 1 mH = 10^-3 H)
τ = L / R = (10.0 x 10^-3 H) / 1000 Ω = 10^-5 s
3. Calculate the voltage drop across the inductor (VL) using the formula:
VL = V * (1 - e^(-t / τ))
V = 30.0 V (given)
t = 20.0 µs = 20.0 x 10^-6 s
Substituting these values into the formula:
VL = 30.0 V * (1 - e^(-20.0 x 10^-6 s / 10^-5 s))
= 30.0 V * (1 - e^(-2))
≈ 30.0 V * (1 - 0.135)
≈ 30.0 V * 0.865
≈ 25.95 V
4. Calculate the voltage drop across the resistor (VR) using Ohm's Law:
VR = I * R
= 0.030 A * 1000 Ω
= 30.0 V
Therefore, the voltage drop across the inductor (VL) 20.0 µs after the switch is closed is approximately 25.95 V, and the voltage drop across the resistor (VR) is 30.0 V.
To find the voltage drops across the resistor (VR) and the inductor (VL) after a given time, we need to apply the principles of electrical circuit analysis.
First, let's determine the current flowing through the circuit. Since the circuit only contains a resistor and an inductor in series, the total impedance (Z) can be calculated using the formula:
Z = R + jωL
Where R is the resistance (1.0 KΩ), ω is the angular frequency, and L is the inductance (10.0 mH).
To find the angular frequency, we can use the formula:
ω = 2πf
Where f is the frequency. Since we don't have the frequency, we need additional information to obtain it.
However, in this case, we can assume that the current flow has reached its steady-state after a long time, so the inductor behaves like a short circuit. This means that the voltage across the inductor VL is zero.
So, to find the voltage drop across the resistor VR:
VR = VL
VL = 0
Therefore, the voltage drops VR and VL would both be zero (0) at t = 20.00 µs after the switch is closed.
Note: If we had the frequency, we could calculate the impedance and find the current using Ohm's Law, and then determine the voltage drops across the resistor and inductor using the current and the respective impedance values.