An initially uncharged 1-mF capacitor has the current
across it. Calculate the voltage across it at t = 2 ms and t = 5 ms.

To find the voltage across a capacitor at specific times when a current is flowing through it, we can use the relationship between current, capacitance, and voltage. The fundamental equation that governs this relationship is given by:

Understanding the Relationship

The current \( I(t) \) flowing into a capacitor is related to the rate of change of voltage \( V(t) \) across it by the formula:

I(t) = C \frac{dV(t)}{dt}

Where:

• I(t) is the current at time \( t \)
• C is the capacitance (in farads)
• V(t) is the voltage across the capacitor at time \( t \)

For a capacitor with a capacitance of 1 mF (or 0.001 F), we can rearrange this equation to find the voltage:

V(t) = \frac{1}{C} \int I(t) dt

Calculating Voltage at Specific Times

To calculate the voltage at \( t = 2 \) ms and \( t = 5 \) ms, we first need to know the expression for the current \( I(t) \). Let's assume a simple case where the current is a constant value, say \( I(t) = 2 \) mA (or 0.002 A) for the sake of this example. This means:

Step 1: Integrate the Current

We can integrate the current over time to find the voltage:

V(t) = \frac{1}{C} \int_0^t I(t) dt

Substituting the values:

V(t) = \frac{1}{0.001} \int_0^t 0.002 dt

Step 2: Evaluate the Integral

The integral of a constant is straightforward:

\int_0^t 0.002 dt = 0.002t

Thus, we have:

V(t) = \frac{1}{0.001} \cdot 0.002t = 2t

Step 3: Calculate Voltage at t = 2 ms

Now, substituting \( t = 2 \) ms (or 0.002 s):

V(0.002) = 2 \cdot 0.002 = 0.004 \text{ V} = 4 \text{ mV}

Step 4: Calculate Voltage at t = 5 ms

Next, substituting \( t = 5 \) ms (or 0.005 s):

V(0.005) = 2 \cdot 0.005 = 0.01 \text{ V} = 10 \text{ mV}

Final Results

In summary, for the assumed constant current of 2 mA:

• At \( t = 2 \) ms, the voltage across the capacitor is 4 mV.
• At \( t = 5 \) ms, the voltage across the capacitor is 10 mV.

Keep in mind that if the current varies with time, you would need the specific function for \( I(t) \) to perform the integration accurately. This example illustrates the basic principles of how to approach the problem using a constant current for simplicity.