Two 6.0 μF capacitors, two 2.2 kΩ resistors, and a 14.5 V source are connected in series. Starting from the uncharged state, how long does it take for the current to drop from its initial value to 1.40 mA
Well, time is money they say! In this case, it's time for some math humor!
Let's use the analogy of a water tank to understand this situation. Imagine you have a tank with a small hole at the bottom, and you're filling it up from a faucet. As the tank fills up, the water pressure increases, causing more water to flow out of the hole.
Similarly, in an RC circuit like this, as the capacitors charge up, the voltage across the resistors decreases, leading to a decrease in current flow. It's like trying to eat a whole pizza by yourself – at first, you eat quickly but slow down as you get full.
Now, to calculate the time it takes for the current to drop to 1.40 mA, we can utilize a formula involving the time constant (τ).
The time constant (τ) is given by the product of the resistance (R) and the capacitance (C) in the circuit. τ = R × C. It represents the time it takes for the current to decrease by approximately 63.2% of its initial value.
So, to find how long it takes for the current to drop to 1.40 mA, we can write the equation: I = I0 × e^(-t/τ), where I is the current at any given time, I0 is the initial current, t is the time passed, and e is the base of the natural logarithm.
We can rearrange the equation to solve for t: t = -τ × ln(I/I0).
Given that the initial current (I0) is known, and you want to find the time (t) when the current (I) drops to 1.40 mA, you can plug in the values of τ, I, and I0 to calculate the time t.
So, grab your calculator and let's get a move on calculating that time!
To find the time it takes for the current to drop from its initial value to 1.40 mA, we need to use the formula for the charging or discharging process of a capacitor in an RC circuit.
The formula for the current in an RC circuit is given by:
I(t) = I0 * e^(-t/RC)
Where:
- I(t) is the current at time t
- I0 is the initial current
- t is the time elapsed
- R is the resistance
- C is the capacitance
In this case, we have two capacitors in series, so the total capacitance (Ct) is the sum of the individual capacitances (C1 and C2):
Ct = C1 + C2
We can calculate the total capacitance:
Ct = 6.0 μF + 6.0 μF = 12.0 μF = 12.0 * 10^(-6) F
Now, we can rearrange the formula to solve for time (t):
t = -RC * ln(I(t)/I0)
Given:
- I0 = initial current = current at t = 0 = 1.40 mA = 1.40 * 10^(-3) A
- R = 2.2 kΩ = 2.2 * 10^(3) Ω
- C = Ct = 12.0 μF = 12.0 * 10^(-6) F
Substituting the given values into the formula:
t = - (2.2 * 10^(3) Ω) * (12.0 * 10^(-6) F) * ln(1.40 * 10^(-3) A / (1.40 mA))
Calculating the value inside the natural logarithm:
t = - (2.2 * 10^(3) Ω) * (12.0 * 10^(-6) F) * ln(0.001)
Now we can simplify:
t = - (2.2 * 10^(3) Ω) * (12.0 * 10^(-6) F) * (-6.9078)
t ≈ 0.182 seconds
Therefore, it takes approximately 0.182 seconds for the current to drop from its initial value to 1.40 mA.
To calculate the time it takes for the current to drop from its initial value to a specific value in a series circuit containing capacitors and resistors, we need to use the concept of time constant.
In a series circuit, the total capacitance (C) and total resistance (R) are determined by the combination of the individual components connected in series. The time constant (τ) is calculated using the formula:
τ = RC
Where:
τ = time constant
R = total resistance
C = total capacitance
First, let's calculate the total resistance (R). Since the two 2.2 kΩ resistors are connected in series, their resistances add up to give us the total resistance:
R = 2.2 kΩ + 2.2 kΩ = 4.4 kΩ
Next, let's calculate the total capacitance (C). Since the two 6.0 μF capacitors are connected in series, their effective capacitance can be calculated using the formula:
1/C = 1/C1 + 1/C2
Where:
C = total capacitance
C1, C2 = individual capacitances
1/C = 1/6.0 μF + 1/6.0 μF = 2/6.0 μF
C = 6.0 μF/2 = 3.0 μF
Now that we have the total resistance (R = 4.4 kΩ) and total capacitance (C = 3.0 μF), we can calculate the time constant (τ):
τ = RC = (4.4 × 10^3 Ω) × (3.0 × 10^-6 F) = 13.2 milliseconds
The time constant represents the time it takes for the current to decrease to approximately 36.8% of its initial value.
To calculate the time it takes for the current to drop from its initial value to 1.40 mA, we can use the formula:
t = - τ × ln(I/I0)
Where:
t = time taken
τ = time constant
I = final current (in amperes)
I0 = initial current (in amperes)
Converting 1.40 mA to amperes:
I = 1.40 mA = 1.40 × 10^-3 A
Using the given initial current (uncharged state), we assume I0 = 0 A.
t = - (13.2 × 10^-3 s) × ln(1.40 × 10^-3 A/0 A)
Since ln(1) = 0, we can simplify the equation further:
t = - (13.2 × 10^-3 s) × ln(1)
Therefore, the time it takes for the current to drop from its initial value to 1.40 mA is approximately 0 seconds, or instantaneously in this case.
RC= 4.4*3.0E-6
I=14.5/4.4 * e^t/RC
take the log (e) of each side.
ln 1.40=ln 14.5 - ln4.4 + t/RC
solve for t.