Question

The total charge that has entered a circuit element is q(t) = 6 (1 - e-7t) when t ≥ 0 and q(t) = 0 when t < 0. The current in the element for t ≥ 0 can be represented as i left-parenthesis t right-parenthesis equals Upper B e Superscript negative a t A where B and a are real constants. Determine the values of B and a.

Answer 1

Answer:

$A=42, B=-7$

Step-by-step explanation:

The current function of time is defined as follows:

$I(t)=\frac{dq(t)}{dt}$

where $q(t)$ is the charge function.

For the given charge function of time $q(t)=6\left( 1-e^{-7t}\right)$ we have the following current function:

$I(t)=\frac{d}{dt} \left(6\left( 1-e^{-7t}\right)\right)=42e^{-7t}$

In the problem it is proposed that $I(t)=Be^{-At}$.

Examining the expression of $I(t)$ we obtained by deriving $q(t)$ with the expression proposed by the problem and comparing term by term:

$I(t)=Be^{-At}=42e^{-7t}$

We conclude that $A=-7$ and $B=42$.