Show that the time for the voltage in the RC circuit to rise to (Vo/2) "half-max" is t1/2 = T ln 2
Time constant T=RC
U =U₀exp(-t/RC)
U/U₀ =exp(-t/RC)
U₀/2U₀ =1/2 = exp(-t/RC)
ln(1/2) = -t/RC)
ln(1/2) =- ln2
ln2 = t/RC
t=RC•ln2=T•ln2
To determine the time for the voltage in an RC circuit to rise to (Vo/2), we need to analyze the charging process and use the exponential behavior of the voltage in the circuit.
The voltage across a charging capacitor in an RC circuit can be described by the equation:
V(t) = Vo * (1 - e^(-t/RC))
Where:
- V(t) is the voltage across the capacitor at time t
- Vo is the initial voltage across the capacitor
- R is the resistance in the circuit
- C is the capacitance in the circuit
- e is the base of the natural logarithm
1. Set V(t) equal to (Vo/2) and solve for t:
(Vo/2) = Vo * (1 - e^(-t/RC))
2. Divide both sides by Vo:
1/2 = 1 - e^(-t/RC)
3. Rearrange the equation:
e^(-t/RC) = 1/2
4. Take the natural logarithm of both sides:
-t/RC = ln(1/2)
5. Multiply both sides by -RC:
t = -RC * ln(1/2)
6. Simplify the equation using the fact that ln(1/2) = -ln(2):
t = RC * ln(2)
7. Finally, rewrite the equation in the canonical form:
t1/2 = T * ln(2)
Where T is the time constant of the RC circuit, given by T = RC.
Therefore, the time for the voltage in the RC circuit to rise to (Vo/2) or "half-max" is t1/2 = T * ln(2).
To show that the time for the voltage in an RC circuit to rise to half of its maximum value, t1/2, is equal to T ln 2, we need to consider the charging behavior of the circuit.
An RC circuit consists of a resistor (R) and a capacitor (C) connected in series. When a voltage source is connected to the circuit, the capacitor charges up gradually.
The voltage across the capacitor (Vc) at any given time can be calculated using the formula:
Vc(t) = Vo(1 - e^(-t/RC))
Where:
- Vc(t) is the voltage across the capacitor at time t
- Vo is the maximum voltage (fully charged voltage)
- e is the mathematical constant Euler's number (approximately 2.71828)
- t is the time elapsed
- R is the resistance
- C is the capacitance
We want to find the time t1/2 it takes for Vc(t) to reach half of its maximum voltage (Vo/2). So we can write the equation:
Vo/2 = Vo(1 - e^(-t1/2/RC))
To find t1/2, we need to isolate it. Divide both sides of the equation by Vo:
1/2 = 1 - e^(-t1/2/RC)
Rearrange the equation:
e^(-t1/2/RC) = 1 - 1/2
Simplify:
e^(-t1/2/RC) = 1/2
Now, take the natural logarithm of both sides of the equation:
ln(e^(-t1/2/RC)) = ln(1/2)
The natural logarithm of e^(-t1/2/RC) simplifies as:
-t1/2/RC = ln(1/2)
Multiply both sides of the equation by -1 to get t1/2 alone:
t1/2 = -RC * ln(1/2)
We can simplify ln(1/2) by using a logarithmic property that states ln(1/x) = -ln(x):
t1/2 = -RC * (-ln(2))
Now, simplify further:
t1/2 = RC * ln(2)
Since a negative sign multiplied by another negative sign cancels out, we finally arrive at:
t1/2 = T ln 2
Where T = RC is the time constant of the RC circuit.
Hence, we have shown that the time for the voltage in the RC circuit to rise to half of its maximum value is T ln 2.