91Ó°ÊÓ

$$\begin{gathered} \mathrm{Current}\mathrm{i}=5\mathrm{A} \\ \mathrm{Time}\mathrm{t}=6\min \\ \mathrm{Emf}\mathrm{Î\mu }=6\mathrm{V} \end{gathered}$$

Write an expression for the change in the energy using the definition of emf. Then, using the definition of current, replace the change in charge in the earlier expression and after inserting the given values, find the reduced chemical energy of the battery

Electromotive forceis defined as the electric potential produced by either electrochemical cell or by changing the magnetic field.

Formulae are as follow:

$$\begin{gathered} \mathrm{Î\mu }=\frac{d\mathrm{W}}{d\mathrm{q}} \\ \mathrm{i}=\frac{∆\mathrm{q}}{∆\mathrm{t}} \end{gathered}$$

Where,

ε is emf, i is current, q is charge, w is work done, t is time.

Emf of the battery is defined as work done per unit charge and is given as,

$$\begin{gathered} \mathrm{Î\mu }=\frac{∆\mathrm{W}}{∆\mathrm{q}} \\ ∆\mathrm{W}=\mathrm{Î\mu }∆\mathrm{q} \end{gathered}$$

5A Current is up in the circuit for time $∆\mathrm{t}=6\min$. Then, the change in work done is nothing but the change in chemical energy $∆\mathrm{W}=∆\mathrm{E}$. Then,

$∆\mathrm{E}=\mathrm{Î\mu }∆\mathrm{q}$

From the definition of current,

$$\begin{gathered} \mathrm{i}=\frac{∆\mathrm{q}}{∆\mathrm{t}} \\ ∆\mathrm{q}=\mathrm{i}∆\mathrm{t} \end{gathered}$$

Hence,

$∆\mathrm{E}=\mathrm{¾\pm Î\mu }∆\mathrm{t}$

Substituting all values in the above equation,

$$\begin{gathered} ∆\mathrm{E}=5\mathrm{A}×6\mathrm{V}×360\mathrm{s} \\ ∆\mathrm{E}=10800~1.1×{10}^4\mathrm{J} \end{gathered}$$

Hence, the chemical energy of the battery reduced is$∆\mathrm{E}=1.1×{10}^4\mathrm{J}$

Therefore, by using the definition of emf and the definition of current the change in energy can be determined.