Current Electricity Question 12
Question 12 - 2024 (29 Jan Shift 1)
The electric current through a wire varies with time as $I=I_{0}+\beta$ t. where $I_{0}=20 \mathrm{~A}$ and $\beta=3 \mathrm{~A} / \mathrm{s}$. The amount of electric charge crossed through a section of the wire in $20 \mathrm{~s}$ is :
(1) $80 \mathrm{C}$
(2) $1000 \mathrm{C}$
(3) $800 \mathrm{C}$
(4) $1600 \mathrm{C}$
Show Answer
Answer: (2)
Solution:
Given that
Current $I=I_{0}+\beta t$
$I_{0}=20 \mathrm{~A}$
$\beta=3 \mathrm{~A} / \mathrm{s}$
$\mathrm{I}=20+3 \mathrm{t}$
$\frac{\mathrm{dq}}{\mathrm{dt}}=20+3 \mathrm{t}$
$\int_{0}^{\mathrm{q}} \mathrm{dq}=\int_{0}^{20}(20+3 \mathrm{t}) \mathrm{dt}$
$q=\int_{0}^{20} 20 d t+\int_{0}^{20} 3 t d t$
$\mathrm{q}=\left[20 \mathrm{t}+\frac{3 \mathrm{t}^{2}}{2}\right]_{0}^{20}=1000 \mathrm{C}$
