Question

The current is increasing at a rate of \(3.60 \mathrm{~A} / \mathrm{s}\) in an \(\mathrm{RL}\) circuit with \(R=3.25 \Omega\) and \(L=440 . \mathrm{mH}\). What is the potential difference across the circuit at the moment when the current in the circuit is \(3.00 \mathrm{~A} ?\)

Short answer

Answer: The potential difference across the RL circuit when the current is 3 A is approximately 11.334 V.

Step-by-step solution

Identify the relevant formula

We will use the formula for the potential difference across an RL circuit which is given by: $$V_t = V_R + V_L$$ where \(V_t\) is the total potential difference, \(V_R\) is the potential difference across the resistor, and \(V_L\) is the potential difference across the inductor.

Calculate the potential difference across the resistor

The potential difference across the resistor can be found using Ohm's law, which is given by: $$V_R = IR$$ where I is the current in the circuit and R is the resistance. Plugging in the given values, we have: $$V_R = (3.00 \mathrm{~A})(3.25 \mathrm{~\Omega}) = 9.75 \mathrm{~V}$$

Calculate the potential difference across the inductor

The potential difference across the inductor can be found using the formula: $$V_L = L\frac{dI}{dt}$$ where L is the inductance and \(\frac{dI}{dt}\) is the rate of change of current. Plugging in the given values, we have: $$V_L = (440 \times 10^{-3} \mathrm{~H})(3.60 \mathrm{~A/s}) = 1.584 \mathrm{~V}$$

Calculate the total potential difference

Now, we can find the total potential difference across the circuit by adding the potential difference across the resistor and the inductor: $$V_t = V_R + V_L = 9.75 \mathrm{~V} + 1.584 \mathrm{~V} = 11.334 \mathrm{~V}$$ So, the potential difference across the circuit at the moment when the current in the circuit is 3 A is approximately 11.334 V.