20 Q 5V 0.25 H In the circuit below, at time t=0 the switch is placed in the position A. What is the initial rate of change in current in the inductor (di/dt)? 20 A/S 10 A/s 5.0 A/S 1.25 A/s 0

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### Educational Content: Initial Rate of Change of Current in Inductor

#### Circuit Description
Below is a diagram of an electrical circuit which includes the following components:
- A 20Ω resistor.
- A 0.25 H (henry) inductor.
- A switch that can be toggled between positions A and B.
- A 5V power supply.

The circuit is arranged in such a way that:
- The resistor and the inductor are connected in series.
- The switch is initially placed at position A.

#### Circuit Diagram
![Circuit Diagram](image link here)
- The circuit consists of a 20Ω resistor.
- There is a 0.25 H inductor in series with the resistor.
- The circuit is powered by a 5V DC supply.
- The switch can be flipped between positions A and B.

#### Problem Statement
At time \( t = 0 \), the switch is placed in the position A. You are asked to determine the initial rate of change of current in the inductor (\( \frac{di}{dt} \)).

##### Options:
- 20 A/s
- 10 A/s
- 5.0 A/s
- 1.25 A/s
- 0

#### Solution Explanation:
To solve this problem, we will use the initial condition after the switch is moved to position A and apply the basic principles of inductance and Ohm's Law.

When the switch is first moved to position A, the inductor initially opposes any change in current. Therefore, the initial voltage across the inductor  \( V_L \) is equal to the battery voltage \( V \).

Using the formula for inductors:

\[ V_L = L \frac{di}{dt} \]

Where:
- \( V_L \) is the voltage across the inductor
- \( L \) is the inductance
- \( \frac{di}{dt} \) is the rate of change of current

Given:
\[ V_L = 5V \]
\[ L = 0.25H \]

Rearrange the formula to solve for \( \frac{di}{dt} \):

\[ \frac{di}{dt} = \frac{V_L}{L} = \frac{5V}{0.25H} = 20 \text{A/s} \]

Thus, the initial rate of change of current in the inductor is **
Transcribed Image Text:### Educational Content: Initial Rate of Change of Current in Inductor #### Circuit Description Below is a diagram of an electrical circuit which includes the following components: - A 20Ω resistor. - A 0.25 H (henry) inductor. - A switch that can be toggled between positions A and B. - A 5V power supply. The circuit is arranged in such a way that: - The resistor and the inductor are connected in series. - The switch is initially placed at position A. #### Circuit Diagram ![Circuit Diagram](image link here) - The circuit consists of a 20Ω resistor. - There is a 0.25 H inductor in series with the resistor. - The circuit is powered by a 5V DC supply. - The switch can be flipped between positions A and B. #### Problem Statement At time \( t = 0 \), the switch is placed in the position A. You are asked to determine the initial rate of change of current in the inductor (\( \frac{di}{dt} \)). ##### Options: - 20 A/s - 10 A/s - 5.0 A/s - 1.25 A/s - 0 #### Solution Explanation: To solve this problem, we will use the initial condition after the switch is moved to position A and apply the basic principles of inductance and Ohm's Law. When the switch is first moved to position A, the inductor initially opposes any change in current. Therefore, the initial voltage across the inductor \( V_L \) is equal to the battery voltage \( V \). Using the formula for inductors: \[ V_L = L \frac{di}{dt} \] Where: - \( V_L \) is the voltage across the inductor - \( L \) is the inductance - \( \frac{di}{dt} \) is the rate of change of current Given: \[ V_L = 5V \] \[ L = 0.25H \] Rearrange the formula to solve for \( \frac{di}{dt} \): \[ \frac{di}{dt} = \frac{V_L}{L} = \frac{5V}{0.25H} = 20 \text{A/s} \] Thus, the initial rate of change of current in the inductor is **
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