MCQ
Current passing through a wire as function of time is given as $I(t)=0.02 t+0.01 \mathrm{~A}$. The charge that will flow through the wire from $t=1 \mathrm{~s}$ to $\mathrm{t}=2 \mathrm{~s}$ is :
- A0.06 C
- B0.02 C
- C0.07 C
- ✓0.04 C
✓
Answer
Correct option: D.
0.04 C
(D) 0.04 C
$\mathrm{q}=\int \mathrm{idt}$
$\int_{0}^{2}(0.02 t+0.01) d t$
$\mathrm{q}=\left[0.02 \frac{\mathrm{t}^{2}}{2}+0.01 \mathrm{t}\right]_{1}^{2}$
$=0.01(3)+0.01(1)$
$=0.04 \mathrm{C}$
$\mathrm{q}=\int \mathrm{idt}$
$\int_{0}^{2}(0.02 t+0.01) d t$
$\mathrm{q}=\left[0.02 \frac{\mathrm{t}^{2}}{2}+0.01 \mathrm{t}\right]_{1}^{2}$
$=0.01(3)+0.01(1)$
$=0.04 \mathrm{C}$
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