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Chapter 12: Problem 1755

Formula for current flowing through a wire is \(I=6 t^{2}+4 t-2\) here \(\mathrm{t}\) is in second and I is an ampere. In this wire, what is the quantity of electric charge passing in time interval \(1 \mathrm{sec}\) to 2 sec? (A) \(8 \mathrm{C}\) (B) \(18 \mathrm{C}\) (C) \(20 \mathrm{C}\) (D) \(24 \mathrm{C}\)

Short Answer

Expert verified
The quantity of electric charge passing through the wire in the time interval from 1 second to 2 seconds is \(18 \mathrm{C}\).

Step by step solution

01

Write down the given formula

The formula for the current flowing through the wire is given as: \[ I(t) = 6t^2 + 4t - 2 \]
02

Integrate the current function with respect to time

To find the electric charge passing through the wire, we need to integrate the current function with respect to the time interval, between 1 second and 2 seconds. The integral can be written as: \[Q = \int_1^2 I(t) dt = \int_1^2 (6t^2 + 4t - 2) dt \]
03

Evaluate the integral

Now, we can evaluate the integral as follows: \[\begin{align*} Q &= \left[ 2t^3 + 2t^2 - 2t \right]_1^2 \\ &= \left[ 2(2)^3 + 2(2)^2 - 2(2) \right] - \left[ 2(1)^3 + 2(1)^2 - 2(1) \right] \\ &= (16 + 8 - 4) - (2 + 2 - 2) \\ &= 20 - 2 \\ &= 18 \mathrm{C} \end{align*}\]
04

State the answer

The quantity of electric charge passing through the wire in the time interval from 1 second to 2 seconds is 18 Coulombs. Thus, the correct answer is (B) 18 C.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Current Function
In the given exercise, we are dealing with a current function that describes how the electrical current flows through a wire. The current function is represented as:
\[ I(t) = 6t^2 + 4t - 2 \] where:
  • \(I(t)\) is the current at time \(t\), measured in amperes (A).
  • \(t\) is the time in seconds.
This formula helps us understand that the current isn't constant; it changes over time as indicated by the quadratic and linear terms. A current function can show how complex and dynamic electrical currents are in real-world circuits.
Integration
To find out how much charge passes through the wire, we integrate the current function over a specific time interval. Integration is a key mathematical tool that allows us to sum up an infinite number of infinitesimally small amounts, which in this case represents the total charge.The integral of the current function \( \int I(t) \, dt \) over the time interval from \( t = 1 \) to \( t = 2 \) is calculated as:
\[ Q = \int_1^2 (6t^2 + 4t - 2) \, dt \]Integration thus helps convert the rate of charge flow (current) into the total charge.
Time Interval
The time interval is crucial when determining how much electric charge flows through the wire. A time interval specifies the start and end point for evaluating the total charge passing through. In this exercise, the given time interval is from 1 second to 2 seconds. It's like saying, "Let's focus only on what's happening within this 1-second slice of time." This constraint simplifies the integration of the current function by giving us clear boundaries.
Coulombs
Coulombs measure the quantity of electricity, representing the charge. One coulomb is defined as the charge transported by a constant current of one ampere in one second.In this problem, after performing the integration, we found that 18 coulombs of charge pass through the wire from \( t = 1 \) to \( t = 2 \). Therefore, simplifying the integral essentially counted up all those tiny little amounts of charge into a single, understandable number: 18 C. This represents the total electric charge that has flowed through the wire during the specified time interval.

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Most popular questions from this chapter

If \(\sigma_{1}, \sigma_{2}\), and \(\sigma_{3}\) are the conductance's of three conductor then equivalent conductance when they are joined in series, will be. (A) \(\sigma_{1}+\sigma_{2}+\sigma_{3}\) (B) \(\left(1 / \sigma_{1}\right)+\left(1 / \sigma_{2}\right)+\left(1 / \sigma_{3}\right)\) (C) \(\left\\{\left(\sigma_{1} \sigma_{2} \sigma_{3}\right) /\left(\sigma_{1}+\sigma_{2}+\sigma_{3}\right)\right\\}\) (D) None of these.

4 cell each of emf \(2 \mathrm{v}\) and internal resistance of \(1 \Omega\) are connected in parallel to a load resistor of \(2 \Omega\) Then the current through the load resistor is.... (A) \(2 \mathrm{~A}\) (B) \(1.5 \mathrm{~A}\) (C) \(1 \mathrm{~A}\) (D) \(0.888 \mathrm{~A}\)

The resistance of the series combination of two resistances is \(\mathrm{S}\), when they are joined in parallel the total resistance is \(\mathrm{P}\) If \(S=n P\), then the minimum possible value of \(n\) is.... (A) 4 (B) 3 (C) 2 (D) 1

A circuit with an infinite no of resistance is shown in fig. the resultant resistance between \(\mathrm{A}\) and \(\mathrm{B}\), when \(\mathrm{R}_{1}=1 \Omega\) and \(\mathrm{R}_{2}=2 \Omega\) will be (A) \(4 \Omega\) (B) \(1 \Omega\) (C) \(2 \Omega\) (D) \(3 \Omega\)

The length of a potentiometer wire is \(600 \mathrm{~cm}\) and it carries a current of \(40 \mathrm{~m} \mathrm{~A}\) for cell of emf \(2 \mathrm{~V}\) and internal resistance \(10 \Omega\), the null point is found to be at \(500 \mathrm{~cm}\) on connecting a voltmeter across the cell, the balancing length is decreased by \(10 \mathrm{~cm}\) The resistance of the voltmeter is (A) \(500 \Omega\) (B) \(290 \Omega\) (C) \(49 \Omega\) (D) \(20 \Omega\)
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