Question

The charge flowing through a resistor R = 2 Q varies with time as q = 3 - 4t². The heat produced in 2 second is about (where q is in C and t is in s)

A)330.3 J

B)341.3 J

C)345.2 J

D)326.4 J​

Answer 1

Answer:

The correct option is B) 341.3 J

Explanation:

Heat H = $I^2Rt$

where I is current, t is time and R is resistance.

Q=It

where Q is charge and t is time.

given,

$q=3-4t^2$

R = 2 Ω

dQ =I dt

I = $\frac{dQ}{dt}$

I = $\frac{d(3-4t^2)}{dt}$

I = -8t

dH = $I^{2} Rdt$

integrating both sides

H= $\int\ {(-8t)^2} R \, dt$

H= $R\int\ {(-8t)^2} \, dt$

H = $\frac {64Rt^3}{3}$

t= 2second and R= 2Ω

H= $\frac{64*2*2^3}{3}$

H = 341.3 J

Hence, heat produced is 341.3 J