If a circuit has a 3.9 k ohms resistor and a 5 uF capacitor, find the current flow in the circuit at 0.005 seconds, if the maximum current flow in the circuit is 1.5 mA.
RC = 3.9k * 5 uF = 19.5 ms = 0.0195 s.
t / RC = 5 ms / 19.5 ms = 0.256 ms =
0.000256 s.
i = i max / e^(t/RC),
i = 1.5 ma / e^0.256 = 1.5 / 1.292 =
1.16 mA @ 5 ms (0.005 s).
s = Seconds.
ms = milliseconds.
To find the current flow in the circuit at a specific time, we need to determine the behavior of the circuit over time. In this case, we have a resistor and a capacitor in the circuit, which means it is an RC circuit.
The behavior of an RC circuit can be described using the equation:
Vc(t) = V0 * (1 - e^(-t/RC))
Where:
- Vc(t) is the voltage across the capacitor at time t
- V0 is the initial voltage across the capacitor (in this case, unknown)
- e is the base of the natural logarithm (approximately 2.71828)
- t is the time (in seconds)
- R is the resistance of the circuit (3.9 k ohms)
- C is the capacitance of the capacitor (5 uF)
We can rearrange the equation to solve for the current flow (I) through the capacitor at a particular time:
I = Vc(t) / R
To find the voltage across the capacitor at a specific time, we need to know the initial voltage. Since it is not given in the question, we cannot provide an exact answer. However, we can still solve the equation using the given values and calculate the maximum current flow in the circuit.
Let's substitute the values into the equation:
I = Vc(t) / R
Vc(t) = V0 * (1 - e^(-t/RC))
I = (V0 * (1 - e^(-t/RC))) / R
Now, with the given values:
- R = 3.9 k ohms (or 3900 ohms)
- C = 5 uF (or 5*10^-6 F)
- t = 0.005 seconds
Let's put these values into the equation and calculate:
I = (V0 * (1 - e^(-0.005/(3900 * 5*10^-6)))) / 3900
I = (V0 * (1 - e^(-0.01))) / 3900
To calculate the maximum current flow in the circuit, you can substitute the maximum allowable voltage across the capacitor (V0) into the equation and solve for I:
1.5 mA = (V0 * (1 - e^(-0.01))) / 3900
Solving this equation requires knowing the initial voltage (V0), which is not provided. Therefore, we cannot determine the exact current at 0.005 seconds without knowing the initial voltage.