Question

Suppose the emf of the battery in the circuit shown in Figure varies with time t so that the current is given by i(t) = 3.0 +5.0 t , where i is in amperes and t is in seconds. Take $\mathit{R}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{0}\mathit{\Omega }\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\mathit{L}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{0}\mathit{H}$, and find an expression for the battery emf as a function of t. (Hint: Apply the loop rule.)

Short answer

Expression for the battery emf as a function of t is$\epsilon =20t+42$

Step-by-step solution

Given

i) Current is$i\left(t\right)=3.0+5.0t$

ii) Resistance is$R=4.0\Omega$

iii) Inductance is$L=6.0H$

iv) Figure 30-16 is the RL circuit.

Understanding the concept

We use the concept of loop rule. Applying loop rule to the circuit, we can write the equation. Plugging the current equation, we can solve for emf.

Formulae:

$\Sigma V=0$

Find an expression for the battery emf as a function of t.

By applying the loop rule, we can write

$$\begin{gathered} \epsilon -L\frac{di}{dt}-iR=0 \\ \epsilon =L\frac{di}{dt}+iR \end{gathered}$$

Plugging the given equation of current, we get

$$\begin{gathered} \epsilon =L\frac{d\left(3.0+5.0t\right)}{dt}+\left(3.0+5.0t\right)R \\ \epsilon =L\left(5.0\right)+\left(3.0+5.0t\right)R \end{gathered}$$

Using the values of inductance and resistance,

$$\begin{gathered} \epsilon =6.0\left(5.0\right)+\left(3.0+5.0t\right)4.0 \\ \epsilon =30+12+20t \\ \epsilon =20t+42 \end{gathered}$$