JEE MAIN - Physics (2021 - 24th February Morning Shift - No. 3)
A current through a wire depends on time as
i = $$\alpha$$0t + $$\beta$$t2
where $$\alpha$$0 = 20 A/s and $$\beta$$ = 8 As$$-$$2. Find the charge crossed through a section of the wire in 15 s.
i = $$\alpha$$0t + $$\beta$$t2
where $$\alpha$$0 = 20 A/s and $$\beta$$ = 8 As$$-$$2. Find the charge crossed through a section of the wire in 15 s.
2250 C
2100 C
260 C
11250 C
Explanation
Given, $$i = {\alpha _0}t + \beta {t^2}$$
where, $$\alpha$$0 = 20 A/s, $$\beta$$ = 8 A/s2
We know that, $$i = {{dq} \over {dt}}$$
$$ \Rightarrow {{dq} \over {dt}} = i = {\alpha _0}t + \beta {t^2} = 20t + 8{t^2}$$
$$ \Rightarrow dq = (20t + 8{t^2})dt$$
On integrating both sides, we get
$$\int\limits_0^q {dq = \int\limits_0^{15} {(2t + 8{t^2})dt} } $$
$$q = \left[ {{{20{t^2}} \over 2} + {{8{t^3}} \over 3}} \right]_0^{15} = 10 \times {(15)^2} + {8 \over 3} \times {(15)^3}$$
$$\therefore$$ q = 11250 C
where, $$\alpha$$0 = 20 A/s, $$\beta$$ = 8 A/s2
We know that, $$i = {{dq} \over {dt}}$$
$$ \Rightarrow {{dq} \over {dt}} = i = {\alpha _0}t + \beta {t^2} = 20t + 8{t^2}$$
$$ \Rightarrow dq = (20t + 8{t^2})dt$$
On integrating both sides, we get
$$\int\limits_0^q {dq = \int\limits_0^{15} {(2t + 8{t^2})dt} } $$
$$q = \left[ {{{20{t^2}} \over 2} + {{8{t^3}} \over 3}} \right]_0^{15} = 10 \times {(15)^2} + {8 \over 3} \times {(15)^3}$$
$$\therefore$$ q = 11250 C
Comments (0)
