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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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Most popular questions from this chapter

In the circuit shown in fig the potential difference across \(3 \Omega\) is. (A) \(2 \mathrm{~V}\) (B) \(4 \mathrm{~V}\) (C) \(8 \mathrm{~V}\) (D) \(16 \mathrm{~V}\)

A Potentiometer wire, \(10 \mathrm{~m}\) long, has a resistance of \(40 \Omega\). It is connected in series with a resistance box and a \(2 \mathrm{v}\) storage cell If the potential gradient along the wire is $0.1 \mathrm{~m} \mathrm{v} / \mathrm{cm}$, the resistance unplugged in the box is. (A) \(260 \Omega\) (B) \(760 \Omega\) (C) \(960 \Omega\) (D) \(1060 \Omega\)

Match the following two columns. Column I \(\quad\) Column II (a) Electrical resistance (p) \(\left[\mathrm{MLT}^{-2} \mathrm{~A}^{2}\right]\) (b) Electric potential (q) \(\left[\mathrm{ML}^{2} \mathrm{~T}^{-3} \mathrm{~A}^{-2}\right]\) (c) Specific resistance (r) \(\left[\mathrm{ML}^{2} \mathrm{~T}^{-3} \mathrm{~A}^{-1}\right]\) (d) Specific conductance (s) None of there (A) \(a-q, b-s, c-r, d-p\) (B) \(a-q, b-r, c-s, d-s\) (C) \(a-p, b-q, c-s, d-r\) (D) \(a-p, b-r, c-q, d-s\)

At what temperature will the resistance of a copper wire be three times its value at \(0^{\circ} \mathrm{C}\) ? (Given: temperature coefficient of resistance for copper \(=4 \times 10^{-3}{ }^{\circ} \mathrm{C}^{-1}\) ) (A) \(400^{\circ} \mathrm{C}\) (B) \(450^{\circ} \mathrm{C}\) (C) \(500^{\circ} \mathrm{C}\) (D) \(550^{\circ} \mathrm{C}\)

How would you arrange 48 cells each of e.m.f \(2 \mathrm{~V}\) and internal resistance \(1.5 \Omega\) so as to pass maximum current through the external resistance of \(2 \Omega\) ? (A) 2 cells in 24 groups (B) 4 cells in 12 groups (C) 8 cells in 6 groups (D) 3 cells in 16 groups
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