Question

A series RL circuit includes a 6.05 V 6.05 V battery, a resistance of R = 0.655 Ω , R=0.655 Ω, and an inductance of L = 2.55 H. L=2.55 H. What is the induced emf 1.43 s 1.43 s after the circuit has been closed?

Answer 1

Answer:

The induced emf 1.43 s after the circuit is closed is 4.19 V

Explanation:

The current equation in LR circuit is :

$I=\frac{V}{R} (1-e^{\frac{-Rt}{L} })$    .....(1)

Here I is current, V is source voltage, R is resistance, L is inductance and t is time.

The induced emf is determine by the equation :

$V_{e}=L\frac{dI}{dt}$

Differentiating equation (1) with respect to time and put in above equation.

$V_{e}= L\times\frac{V}{R}\times\frac{R}{L}e^{\frac{-Rt}{L} }$

$V_{e}=Ve^{\frac{-Rt}{L} }$

Substitute 6.05 volts for V, 0.655 Ω for R, 2.55 H for L and 1.43 s for t in the above equation.

$V_{e}=6.05e^{\frac{-0.655\times1.43}{2.55} }$

$V_{e}=4.19\ V$

Answer 2

Based on the calculations, the induced emf is equal to 4.19 Volts.

Given the following data:

• Voltage = 6.05 V.

• Resistance = 0.655 Ω.

• Inductance = 2.55 H.

• Time = 1.43 seconds.

How to determine the induced emf.

In a RL circuit, current is given by this mathematical expression:

$I=\frac{V}{R} (1-e^{\frac{Rt}{L} })$

Where:

• I is the current.

• V is the source voltage.

• R is the resistance.

• L is the inductance.

• t is the time.

For an induced emf in a circuit, we have:

$E=L\frac{dI}{dt} \\\\E=L \times I \\\\E=L \times \frac{V}{R} (1-e^{\frac{-Rt}{L} })\\\\E=V e^{\frac{-Rt}{L} }\\\\E=6.05 e^{\frac{-0.655 \times 1.43}{2.55} }$

E = 4.19 Volts.

Read more current here: https://brainly.com/question/25813707