A circuit is composed of a capacitor C= 3 micro F, two resistors both with resistance 69 Ohm, an inductor 6.00e-2 H, a switch , and a battery 25 V. The internal resistance of the battery can be ignored . Reminder: The " L=x.xxenn H" notation means " L=x.xx X 10^nn H".
Initially, the switch S is open as in the figure above and there is no charge on the capacitor and no current C 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)?
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What is the current through the inductor (in Amps) at t=8.70e-4 s?
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What is the current through the inductor (in Amps) a long time later?
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What is the current through the capacitor (in Amps) at t= 0 (i.e. just after the switch is closed)?
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What is the current through the capacitor (in Amps) at t= 2.07e-4 s?
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What is the current through the capacitor (in Amps) a long time later?
last ques radially outward both and last ques =0
a)0
b)---
c)V/R=25/69
d)V/R=25/69
e)(v/R)*exp(-t/(R*C)=(25/69)*exp(-t/(69*3e-6)
f)0
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5,7,9,10,11
To find the current through the components in the circuit, we need to analyze the behavior of the circuit after the switch is closed. This is a basic RL circuit with a capacitor in parallel to the inductor.
Let's start by analyzing the behavior just after the switch is closed, at t=0. Since there is no charge on the capacitor and no current flowing through the inductor initially, the current through the inductor (IL) is initially zero.
To find the current through the inductor at t=8.70e-4 s, we can use the formula for the current in a charging RL circuit:
IL(t) = (V/R) * (1 - e^(-t/(L/R)))
Where:
V is the battery voltage (25 V)
R is the resistance (69 Ω)
L is the inductance (6.00e-2 H)
t is the time
Plugging in the given values, we can calculate the current at that specific time.
For the current through the inductor a long time later, we assume that the circuit has reached a steady state. In a steady state, the capacitor acts as an open circuit and no current flows through it. Therefore, the current through the inductor is essentially determined by the resistance and inductance values only.
Since the battery has no internal resistance, the current through the inductor can be calculated using Ohm's Law:
IL = V / R
For the current through the capacitor, we can apply the same logic. Just after the switch is closed (t=0), there is no charge on the capacitor, so the voltage across the capacitor (VC) is initially zero. Therefore, the current through the capacitor (IC) is also initially zero.
To find the current through the capacitor at t=2.07e-4 s, we can use the formula for the current in a charging RC circuit:
IC(t) = (V/R) * e^(-t/(RC))
Where:
V is the battery voltage (25 V)
R is the resistance (69 Ω)
C is the capacitance (3 microF = 3e-6 F)
t is the time
Plugging in the given values, we can calculate the current at that specific time.
For the current through the capacitor a long time later, we assume that the circuit has reached a steady state. In a steady state, the capacitor acts as a short circuit and the current through it is essentially determined by the resistance and voltage values only.
Therefore, the current through the capacitor can be calculated using Ohm's Law:
IC = V / R