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Chapter 14: Problem 2047

The time constant of a LR circuit is \(80 \mathrm{~ms}\), The circuit is connected at \(t=0\) and the steady state current is found to be $4 \mathrm{~A}\(. Find the current at \)20 \mathrm{~ms}$. (a) \(0.98 \mathrm{~A}\) (b) \(1 \mathrm{~A}\) (c) \(0.44 \mathrm{~A}\) (d) \(0.88 \mathrm{~A}\)

Short Answer

Expert verified
The current at \(20\mathrm{~ms}\) in the given LR circuit is approximately \(0.88\mathrm{~A}\).

Step by step solution

01

Understand the LR circuit formula

An LR circuit is made up of an inductor (L) and a resistor (R) connected in series. The current through the circuit grows exponentially when a voltage is applied. We can use the formula to describe the current in an LR circuit at any given time: \[I(t) = I_\infty (1 - e^{- \frac{t}{\tau}})\] where \(I(t)\) is the current at time \(t\), \(I_\infty\) is the steady-state current, \(t\) is the time, and \(\tau\) is the time constant of the LR circuit.
02

Plug in the given values

We are given the time constant, τ = \(80\mathrm{~ms}\), the steady-state current, \(I_\infty = 4\mathrm{~A}\), and the time we want to find the current at, \(t = 20\mathrm{~ms}\). Plug these values into the formula: \[I(20\mathrm{~ms}) = 4\mathrm{~A} (1 - e^{- \frac{20\mathrm{~ms}}{80\mathrm{~ms}}})\]
03

Simplify the expression

Now simplify the expression by dividing the time \(t\) by the time constant τ: \[I(20\mathrm{~ms}) = 4\mathrm{~A} (1 - e^{- \frac{1}{4}})\]
04

Calculate the current at 20 ms

Calculate the value of the exponential term and then multiply by the steady-state current: \[I(20\mathrm{~ms}) = 4\mathrm{~A} (1 - 0.7788)\] \[I(20\mathrm{{~ms}) = 4\mathrm{~A}(0.2212)\] \[I(20\mathrm{~ms}) = 0.8848\mathrm{~A}\]
05

Identify the closest answer

Looking at the given answer choices, the closest answer to our calculated value is: (d) \(0.88\mathrm{~A}\)

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Most popular questions from this chapter

A coil having n turns \(\&\) resistance \(R \Omega\) is connected with a galvanometer of resistance \(4 \mathrm{R} \Omega\). This combination is moved from a magnetic field \(\mathrm{W}_{1} \mathrm{~Wb}\) to $\mathrm{W}_{2} \mathrm{~Wb}\( in \)\mathrm{t}$ second. The induced current in the circuit is.... (a) $-\left[\left\\{\mathrm{W}_{2}-\mathrm{W}_{1}\right\\} /\\{5 \mathrm{Rnt}\\}\right]$ (b) $-\mathrm{n}\left[\left\\{\mathrm{W}_{2}-\mathrm{W}_{1}\right\\} /\\{5 \mathrm{Rt}\\}\right]$ (c) $-\left[\left\\{\mathrm{W}_{2}-\mathrm{W}_{1}\right\\} /\\{\mathrm{Rnt}\\}\right]$ (d) $-\mathrm{n}\left[\left\\{\mathrm{W}_{2}-\mathrm{W}_{1}\right\\} /\\{\mathrm{Rt}\\}\right]$

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