Question

A 5μF capacitor is charged to 30 V and is then connected in series with a 10μH inductor and a 50Ω resistor. The initial current in the circuit will be_____

Answer 1

**Question: A 5μF capacitor is charged to 30 V and is then connected in series with a 10μH inductor and a 50Ω resistor. The initial current in the circuit will be_____**

Final Answer:

When a charged capacitor is connected in series with a resistor and inductor, the initial current is given by
`\(i(t) = (V_0)/(R) \cdot e^{-(t)/(RC)}\).` Substituting the given values, the initial current is 0.6 A. Thus, the circuit's initial state involves a current flow of 0.6 A as the capacitor discharges. The initial current in the circuit will be zero.

Step-by-step explanation:

When a charged capacitor is connected in series with an inductor and a resistor, the initial current in the circuit is determined by the discharge of the capacitor. The formula for the current (i) in an RC circuit is given by
`\(i(t) = (V_0)/(R) \cdot e^{-(t)/(RC)}\)`, where V₀ is the initial voltage across the capacitor, R is the resistance, C is the capacitance, and t is time. In this case, the initial voltage (V₀) is 30 V, the resistance (R) is 50 Ω, and the capacitance (C) is 5μF.

As the capacitor discharges, the exponential term approaches zero, and the current gradually decreases. At the initial moment of connection, t =0, the exponential term becomes
`\(e^0 = 1\)`, and the current is given by
`\(i(0) = (30)/(50) \cdot 1 = 0.6\)` A. Therefore, the initial current in the circuit is 0.6 A.

In summary, the initial current in the circuit is determined by the discharge of the capacitor, and in this case, it is 0.6 A.

Answer 2

The initial current in the circuit is approximately 7.5 μA.

To find the initial current in the circuit, we can use the formula for the transient current in an RLC circuit:

`\[I(t) = \frac{V}{\sqrt{R^2 + \left((1)/(\omega L) - \omega C\right)^2}}\]`

Where:

- I(t) is the current at time t.

- V is the initial voltage across the capacitor.

- R is the resistance (50Ω).

- L is the inductance (10μH).

- C is the capacitance (5μF).

- ω is the angular frequency, given by ω = 1/√(LC).

First, calculate ω:

`\[ω = \frac{1}{\sqrt{(10×10^(-6) H) × (5×10^(-6) F)}} = 1,000,000 rad/s\]`

Now, plug the values into the formula:

`\[I(t) = \frac{30 V}{\sqrt{(50Ω)^2 + \left((1)/(1,000,000 rad/s × 10×10^(-6) H) - 1,000,000 rad/s × 5×10^(-6) F\right)^2}}\]`

Simplify:

`\[I(t) = \frac{30 V}{\sqrt{50Ω^2 + (1 - 5)^2 * 10^(12)}} = \frac{30 V}{\sqrt{50Ω^2 + 16 * 10^(12)}} \approx (30 V)/(4 * 10^(6)) = 7.5 * 10^(-6) A = 7.5 μA\]`

So, the initial current in the circuit is approximately 7.5 μA.