Question

For the given equation of charge Q=3t²+6t, where Q is in coulombs and t is time in seconds, the initial value of current is:

• A. 0 A
• B. 2 A
• C. 6 A
• D. 9 A

Answer 1

The Correct Option is C

The charge equation is given by:

\[ Q(t) = 3t^2 + 6t \]

where:

• \( Q \) represents charge in coulombs (C)
• \( t \) represents time in seconds (s)

To find the initial current.

Current (\( I \)) is the time derivative of charge:

\[ I(t) = \frac{dQ}{dt} \]

Differentiating the charge equation yields:

\[ \frac{dQ}{dt} = \frac{d}{dt}(3t^2 + 6t) \]

\[ I(t) = 6t + 6 \]

The initial current occurs at \( t = 0 \):

\[ I(0) = 6(0) + 6 = 6 \text{ A} \]

The initial current value is \({6 \text{ A}}\).