** i = 120 mA, t<= 0 **

\[ \boldsymbol{ i(t) \ = \ A_1e^{ -500t } \ + \ A_2e^{ -2000t } \ A, \ t \ge 0 } \]

**The potential difference across inductor terminals is 3V at time t = 0.**

**Calculate the mathematical formula of the voltage for time t > 0.****Calculate the time at which the inductor stored power decays to zero.**

The aim of this question is to understand the **current and voltage relationship** of an **inductor** element.

To solve the given question we will use the **mathematical form** of the inductor **voltage-current relationship**:

\[ v(t) = L \dfrac{ di(t) }{ dt } \]

where, $L$ is the **inductance** of the inductor coil.

## Expert Answer

**Part (a): Calculating the equation of voltage across the inductor.**

**Given:**

\[ i(t) \ = \ A_1e^{ -500t } \ + \ A_2e^{ -2000t } \]

**At $ t \ = \ 0 $ :**

\[ i(0) \ = \ A_1e^{ -500(0) } \ + \ A_2e^{ -2000(0) } \]

\[ i(0) \ = \ A_1 \ + \ A_2 \]

**Substituting $ i(0) \ = \ 120 \ = \ 0.12 $ in above equation:**

\[ A_1 \ + \ A_2 \ = \ 0.12 \ … \ … \ … \ (1) \]

**Voltage of an inductor** is given by:

\[ v(t) = L \dfrac{ di(t) }{ dt } \]

**Substituting** value of $ i(t) $

\[ v(t) = L \dfrac{ d }{ dt } \bigg ( A_1e^{ -500t } \ + \ A_2e^{ -2000t } \bigg ) \]

\[ v(t) = L \bigg ( -500A_1e^{ -500t } \ – \ 2000A_2e^{ -2000t } \bigg ) \]

\[ v(t) = ( 50 \times 10^{ -3 } ) \bigg ( -500A_1e^{ -500t } \ – \ 2000A_2e^{ -2000t } \bigg ) \]

\[ v(t) = -25A_1e^{ -500t } \ – \ 100A_2e^{ -2000t } \ … \ … \ … \ (2) \]

At $ t \ = \ 0 $ :

\[ v(0) = -25A_1e^{ -500( 0 ) } \ – \ 100A_2e^{ -2000( 0 ) } \]

\[ v(0) = -25A_1 \ – \ 100A_2 \]

Since, $ v(0) = 3 $, above equation becomes :

\[ -25A_1 \ – \ 100A_2 = 3 \ … \ … \ … \ (3) \]

**Solving equations** $1$ and $3$ simultaneously:

\[ A_1 = 0.2 \ and \ A_2 = -0.08 \]

**Substituting** these values in equation $2$:

\[ v(t) = -25(0.2)e^{ -500t } \ – \ 100(-0.08)e^{ -2000t } \]

\[ v(t) = -5e^{ -500t } \ + \ 8e^{ -2000t } \ V \]

**Part(b): Calculating the time when the energy in the inductor becomes zero.**

Given:

\[ i(t) \ = \ A_1e^{ -500t } \ + \ A_2e^{ -2000t } \]

**Substituting** values of constants:

\[ i(t) \ = \ 0.2 e^{ -500t } \ – \ 0.08 e^{ -2000t } \]

Energy is zero when the **current becomes zero**, so under the given condition:

\[ 0 \ = \ 0.2 e^{ -500t } \ – \ 0.08 e^{ -2000t } \]

\[ \Rightarrow 0.08 e^{ -2000t } \ = \ 0.2 e^{ -500t } \]

\[ \Rightarrow \dfrac{ e^{ e^{ -500t } }{ -2000t } } \ = \ \dfrac{ 0.08 }{ 0.2 } \]

\[ \Rightarrow e^{ 1500t } \ = \ 0.4 \]

\[ \Rightarrow 1500t \ = \ ln( 0.4 ) \]

\[ \Rightarrow t \ = \ \dfrac{ ln( 0.4 ) }{ 1500 } \]

\[ \Rightarrow t \ = \ -6.1 \times 10^{-4} \]

**Negative time** means that there is a **continuous source of energy connected** to the inductor and there is **no plausible time** when the power becomes zero.

## Numerical Result

\[ v(t) = -5e^{ -500t } \ + \ 8e^{ -2000t } \ V \]

\[ t \ = \ -6.1 \times 10^{-4} s\]

## Example

Given the following current equation, find the equation for the voltage for an inductor of inductance $ 1 \ H $:

\[ i(t) = sin(t) \]

Voltage of an inductor is given by:

\[ v(t) = L \dfrac{ di(t) }{ dt } \]

\[ \Rightarrow v(t) = (1) \dfrac{ d }{ dt } ( sin(t) ) \]

\[ \Rightarrow v(t) = cos(t) \]