A circuit is composed of a capacitor C=3 μF, two resistors both with resistance R=69 Ohm, an inductor L=6.00e-2 H, a switch S, and a battery V=25 V. The internal resistance of the battery can be ignored . Reminder: The "L=x.xxenn H" notation means "L=x.xx×10nn H".
(The battery is parallel with a inductor with a resitance in serie and is parallel with a capacitor in serie with a resistance).
Initially, the switch S is open as in the figure above and there is no charge on the capacitor C and no current flowing through the inductor L. At t=0 we close the switch.
Define the current through the inductor to be positive if it flows through the inductor and then through the resistor and therefore down in the drawing. Similarly, define the current through the capacitor to be positive if it flows down in the drawing.
What is the current through the inductor (in Amps) at t=0 (i.e. just after the switch is closed)?
Initially, the switch S is open as in the figure above and there is no charge on the capacitor C and no current flowing through the inductor L. At t=0 we close the switch.
Define the current through the inductor to be positive if it flows through the inductor and then through the resistor and therefore down in the drawing. Similarly, define the current through the capacitor to be positive if it flows down in the drawing.
What is the current through the inductor (in Amps) at t=0 (i.e. just after the switch is closed)?Initially, the switch S is open as in the figure above and there is no charge on the capacitor C and no current flowing through the inductor L. At t=0 we close the switch.
Define the current through the inductor to be positive if it flows through the inductor and then through the resistor and therefore down in the drawing. Similarly, define the current through the capacitor to be positive if it flows down in the drawing.
What is the current through the inductor (in Amps) at t=0 (i.e. just after the switch is closed)?Initially, the switch S is open as in the figure above and there is no charge on the capacitor C and no current flowing through the inductor L. At t=0 we close the switch.
Define the current through the inductor to be positive if it flows through the inductor and then through the resistor and therefore down in the drawing. Similarly, define the current through the capacitor to be positive if it flows down in the drawing.
What is the current through the inductor (in Amps) at t=0 (i.e. just after the switch is closed)?
What is the current through the inductor (in Amps) at t=8.70e-4 s?
What is the current through the inductor (in Amps) a long time later?
What is the current through the capacitor (in Amps) at t=0 (i.e. just after the switch is closed)?
What is the current through the capacitor (in Amps) at t=2.07e-4 s?
What is the current through the capacitor (in Amps) a long time later?
a)0
b)---
c)V/R=5/11
d)V/R=5/11
e)(v/R)*exp(-t/(R*C)=(5/11)*exp(-t/(11*2e-6)
f)0
help me in
5a,9,10,11
To find the current through the inductor and capacitor at different times, we need to analyze the circuit and use the principles of electrical circuits. Let's break down each question and find the respective answers step by step:
1. What is the current through the inductor (in Amps) at t=0?
To find the current through the inductor just after the switch is closed, we need to analyze the circuit in its initial state. Since the switch is open, there is no current flowing through the inductor or the capacitor at t=0.
Therefore, the current through the inductor at t=0 is 0 Amps.
2. What is the current through the inductor (in Amps) at t=8.70e-4 s?
To find the current through the inductor at t=8.70e-4 s, we need to consider the behavior of the circuit after the switch is closed.
When the switch is closed, the capacitor will begin to charge and the current will start flowing through the circuit. We can use the concepts of RL (resistor-inductor) circuits to analyze the current through the inductor.
First, we calculate the time constant (τ) of the RL circuit:
τ = L / R = 6.00e-2 H / 69 Ω = 8.70e-4 s
At t = τ, the current through the inductor will reach approximately 63.2% of its maximum value.
Therefore, the current through the inductor at t=8.70e-4 s will be approximately: 0.632 * (peak current through inductor).
3. What is the current through the inductor a long time later?
A long time later, the capacitor will be fully charged, and the current through the inductor will approach zero. This is because the energy stored in the capacitor will eventually be fully transferred to the inductor, causing the current to decay.
As the current approaches zero, it becomes negligible and can be considered as zero.
Therefore, the current through the inductor a long time later is 0 Amps.
4. What is the current through the capacitor (in Amps) at t=0?
Similar to the current through the inductor at t=0, the current through the capacitor is also 0 Amps at t=0 since the switch is open and no current is flowing through the circuit.
5. What is the current through the capacitor (in Amps) at t=2.07e-4 s?
To find the current through the capacitor at t=2.07e-4 s, we need to consider the behavior of the circuit after the switch is closed.
When the switch is closed, the capacitor starts to charge, and the current starts flowing through the circuit. Using the concepts of RC (resistor-capacitor) circuits, we can analyze the current through the capacitor.
First, we calculate the time constant (τ) of the RC circuit:
τ = R * C = 69 Ω * 3 μF = 69e-6 s
At t = τ, the current through the capacitor will reach approximately 63.2% of its maximum value.
Therefore, the current through the capacitor at t=2.07e-4 s will be approximately: 0.632 * (peak current through capacitor).
6. What is the current through the capacitor (in Amps) a long time later?
A long time later, the capacitor will be fully charged, and the current through the capacitor will approach zero. This is because no current can flow through a fully charged capacitor.
Therefore, the current through the capacitor a long time later is 0 Amps.