Sankalp JEE Full Test-4 Question-27 Solution

Question 27 The electric current through a wire varies with time as I=I_{0}+\beta t.
Here, I_{0}=20 A \beta=3 A/s The amount of electric charge crossing a section of the wire in 20 s is :

(1) 80 C

(2) 1000 C

(3) 800 C

(4) 1600 C

Answer: (2)

Explanation:

The current is given by I=I_{0}+\beta t

Substituting values, I = 203t

Electric current is the rate of flow of charge,

\frac{dq}{dt}=20+3t

Integrate with respect to time from

t=0 to t=20

\int_{0}^{q}dq=\int_{0}^{20}(20+3t)dt

Split the integration,

q=\int_{0}^{20}20\,dt+\int_{0}^{20}3t\,dt q=\left[20t+\frac{3t^{2}}{2}\right]_{0}^{20} q=400+600

q=1000 C