Current Electricity

JEE Main level, single correct, medium

29 of 380

Question 29 of 380

A current through a wire depends on time as i=α0t+βt2i={\alpha }_{0}t+\beta {t}^{2}, where α0=20As−1{\alpha }_{0}=20A{s}^{-1} and β=8As−2\beta =8A{s}^{-2}. Find the charge crossed through a section of the wire in 15s.15s.

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Answer: (B) 11250C11250C

i=20t+8t2i=20t+8{t}^{2}
i=dqdt⇒∫dq=∫idti=\dfrac{dq}{dt}\Rightarrow \int dq=\int idt
⇒q=∫015(20t+8t2)dt\Rightarrow q={\int }_{0}^{15}\left(20t+8{t}^{2}\right)dt
q=(20t22+8t33)015q={\left(\dfrac{20{t}^{2}}{2}+\dfrac{8{t}^{3}}{3}\right)}_{0}^{15}
q=10×(15)2+8(15)33q=10\times {\left(15\right)}^{2}+\dfrac{8{\left(15\right)}^{3}}{3}
q=2250+9000q=2250+9000
q=11250Cq=11250C

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