Charge Q coulombs at time t seconds is given by the differential equation:
0
C
Q
dt
dQ
R where C is the capacitance in farads and R is the resistance in
ohms. Solve the equation for Q given that Q=Q0 when t=0.
A circuit possesses a resistance of 250kΩ and a capacitance of 8.5μF, and
after 0.32 seconds the charge falls to 8.0C. Determine the initial charge and the
charge after 1 second, each correct to 3 significant figures.
T=RC = 250K * 8.5uF = 2125ms = 2.125s.
t/T = 0.32S / 2.125S = 0.1506.
Q = Qo / e^(t/T) = 8.
Qo / e^(0.1506) = 8,
Cross multiply:
Qo = 8*e^(0.1506),
Qo = 8 * 1.1625 = 9.300C. @ t = 0.
t/T = 1S / 2.125s = 0.4706.
Q = Qo / e^(t/T),
Q = 9.30 / e^(0.4706),
Q = 9.30 / 1.60 = 5.809C after 1sec.
t/T=1S /2.125s=0.4706.
To solve the differential equation and determine the initial charge and charge after 1 second, we need to integrate the equation with the given values for resistance (R) and capacitance (C).
The given differential equation is:
dQ/dt = Q/(RC)
Let's first find the value of RC by substituting the given values:
RC = (8.5μF)(250kΩ)
= (8.5 × 10^-6 F)(250 × 10^3 Ω)
= 2.125 ΩF
Now, let's solve the differential equation using separation of variables and integrate both sides:
∫(1/Q)dQ = ∫(1/RC)dt
ln|Q| = (1/RC)t + C1
To simplify, let's take the exponential of both sides:
|Q| = e^((1/RC)t + C1) = e^(1/RC)t * e^C1
Since Q is a positive quantity, we can rewrite the equation as:
Q = ±e^(1/RC)t * C2
Now, let's find the initial charge Q0 when t = 0:
Q0 = ±e^(1/RC)(0) * C2 = ±C2
Since the charge cannot be negative, we can rewrite it as:
Q0 = C2
The initial charge Q0 is equal to C2.
Next, let's find the charge after 1 second:
Q1 = ±e^(1/RC)(1) * C2 = ±e^(1/2.125)(1) * Q0
Since e^(1/2.125) is a constant value, we can simplify it further:
Q1 = k * Q0
To find the value of k, we can substitute the given values:
Q1 = (8.0C) / e^(1/2.125)(0.32s) * Q0
Now we can calculate the values for Q0 and Q1 using the given values:
Q0 = C2 = Q1 / (8.0C) * e^(1/2.125)(0.32s)
Using a calculator, evaluate the value of Q0 to 3 significant figures.
To find Q1 after 1 second, substitute t = 1 second into the equation and evaluate the value using a calculator.
Keep in mind that the final answer will have a positive sign since the given charge falls to 8.0C.
To solve the given differential equation, we need to separate the variables and integrate both sides. Let's start by rewriting the differential equation:
dQ/dt = Q/RC
Now, we can separate the variables by moving dt to the right side and Q/RC to the left side:
dQ/Q = dt/RC
Integrating both sides:
∫(1/Q) dQ = ∫(1/RC) dt
After integrating, we get:
ln|Q| = (1/RC) t + C1
Where C1 is the constant of integration.
To determine the constant of integration, we can use the initial condition given in the problem: Q = Q0 when t = 0. Plugging these values in:
ln|Q0| = (1/RC) * 0 + C1
C1 = ln|Q0|
Now, substituting the value of C1 back into the equation:
ln|Q| = (1/RC) t + ln|Q0|
We can simplify this equation using logarithmic properties:
ln|Q| = ln(|Q0|) + (1/RC) t
Since ln|Q| and ln(|Q0|) are both natural logarithms, we can combine them:
ln(|Q|/|Q0|) = (1/RC) t
Next, we will solve for Q by exponentiating both sides:
|Q|/|Q0| = e^[(1/RC) t]
Since the absolute value doesn't affect our final answer, we can remove it:
Q/|Q0| = e^[(1/RC) t]
To find the particular solution, we need to determine the value of |Q0|. We can use the given information that after 0.32 seconds, the charge falls to 8.0 C:
8.0/|Q0| = e^[(1/RC) * 0.32]
Solving for |Q0|:
|Q0| = 8.0 / e^[(1/RC) * 0.32]
Now we can substitute the values of R and C given in the problem:
|Q0| = 8.0 / e^[(1/(250kΩ * 8.5μF)) * 0.32]
Calculating this expression will give us the value of |Q0|. After finding |Q0|, we can use it to determine the charge after 1 second by substituting t=1 into our equation:
Q = |Q0| * e^[(1/RC) * 1]
Calculating this expression will give us the charge after 1 second. Remember to round the values to 3 significant figures, as stated in the problem.