Emf Of A Cell

What is a Cell?

A cell, in the context of electricity, is a device that converts chemical energy into electrical energy. It consists of two electrodes (an anode and a cathode) immersed in an electrolyte, a substance that conducts electricity. When the cell is connected to a circuit, a chemical reaction occurs between the electrodes and the electrolyte, which generates an electric current. This current flows through the circuit, powering electrical devices.

The Direction of Flow of Current

1. inside the cell is from negative to positive electrode

2. while outside the cell is from positive to negative electrode.

Electromotive Force (Emf) of a Cell

It is the work done / energy carried by a unit charge passing through one complete cycle of the circuit.

E=W/q

unit of emf is volt.

The EMF of a cell is also known as the potential difference across the terminals of a cell when it does not give any current.

Note- Cell is a source of constant emf but not constant current.

Potential Difference

It is the work done / energy carried by the unit charge passing through the external part (excluding the cell ) of the circuit.

The potential difference is equal to the product of the current and the resistance of that given part.

i.e. $V_{A B}=i R$.

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Solved Examples Based on Cell and Emf of a Cell

Example 1: In the circuit shown, current (in A) through the 50 V and 30 V batteries are, respectively

1) 2.5 and 3

2) 3.5 and 2

3) 4.5 and 1

4) 3 and 2.5

Solution:

KVL in loop abgha
$
\begin{aligned}
20 \mathrm{I}_1 & =50 \\
\mathrm{I}_1 & =2.5 \mathrm{~A}
\end{aligned}
$

KVL in loop abcdefgha
$
\begin{aligned}
& 50-5 I_2-30-5 I_2=0 \\
& I_2=2 \mathrm{~A}
\end{aligned}
$

KVL in loop cdefc
$
\begin{aligned}
& 30=10\left(\mathrm{I}_2+\mathrm{I}_3\right) \\
& \Rightarrow \mathrm{I}_2+\mathrm{I}_3=3 \\
& \mathrm{I}_3=3-2=1 \mathrm{~A}
\end{aligned}
$

therefore Current through 50 V battery is $=\mathrm{I}_1+\mathrm{I}_2=2.5+2.0=4.5 \mathrm{~A}$ current through 30 V battery $=\mathrm{I}_3=1 \mathrm{~A}$

Hence, the answer is the option (3).

Example 2:

In the above circuit value of R=10 r. If a total current of 1A is flowing through the circuit the value of E is?

1) 10 r
2) $11 r$
3) $9 r$
4) 0

Solution:

Electromotive force (Emf) of a cell

It is the work done / energy carried by a unit charge passing through one complete cycle of the circuit.

E=W/q

unit of emf is volt.

$\begin{aligned} & \frac{E}{10 r+r}=1 \\ & E=11 r\end{aligned}$

Hence, the answer is the option (2).

Example 3: A cell E₁ of emf 6V and internal resistance $2 \Omega$ is connected with another cell E₂ of emf 4V and internal resistance $8 \Omega$ ( as shown in the figure). The potential difference across points X and Y is:

1) 3.6V

2) 5.6V

3) 10.0 V

4) 2.0V

Solution:

The figure given in the question can be redrawn as given below

The direction of current in the circuit will be as shown in the figure.
So point Y is at a higher potential than X.
So $\mathrm{V}_{\mathrm{Y}}>\mathrm{V}_{\mathrm{X}}$
Current(I) in circuit,
$
\mathrm{I}=\frac{\mathrm{E}_1+\mathrm{E}_2}{\mathrm{r}_1+\mathrm{r}_2}=\frac{(6-4) \mathrm{V}}{(2+8) \Omega}=0.2 \mathrm{amp}
$

For positive potential $A$ is near to positive terminal of $E_2$ so has +4 V$
So potential across $\mathrm{E}_1$ and $\mathbf{E}_2$
$
\begin{aligned}
& \mathrm{E}_1=\mathrm{V}-\mathrm{Ir}_1=6-0.2 \times 2=6-0.4=5.6 \mathrm{~V} \\
& \mathrm{E}_2=\mathrm{V}+\mathrm{Ir}_2=4+0.2 \times 8=4+1.6=5.6 \mathrm{~V}
\end{aligned}
$

So potential between X and $\mathrm{Y}=E_2=5.6$ Volt.

Hence, the answer is the option (2).

Example 4:

In the above circuit value of R= 10 ohm and r= 1 ohm. If a total current of 1A is flowing through the circuit the potential drop (in V) across R is?

1) 10

2) 9

3) 11

4) 0

Solution:

Potential difference (V) It is the work done / energy carried by unit charge passing through the external part (excluding the cell ) of the circuit.

The potential difference is equal to the product of the current and the resistance of that given part.

i.e. $V_{A B}=i R$.
$
V=I R=10 \mathrm{~V}
$

Hence, the answer is the option (1).

Example 5: A voltmeter of resistance $1000 \Omega$ is connected across resistance $500 \Omega$ in a given circuit. What will be the reading of the voltmeter?

1) 4

2) 2

3) 6

4) 1

Solution:

$
\begin{aligned}
R_{e q} & =\frac{1000 * 500}{1500}+500 \\
& =500+\frac{1000}{3}=\frac{2500}{3} \Omega \\
I & =\frac{10}{\left(\frac{2500}{3}\right)} A=\frac{3}{250} A
\end{aligned}
$

Potential Difference across Voltmeter
$
\begin{aligned}
& \mathrm{V}=\mathbb{R} \\
& I_v=\left(\frac{3}{250}\right) \cdot \frac{R_2}{R_1+R_2}=\left(\frac{3}{250}\right) * \frac{500}{1500} \\
& I_v=\frac{1}{250} \mathrm{~A} \\
& \text { Potential Difference }=\frac{1}{250} * 1000 \mathrm{~V}=4 \mathrm{~V}
\end{aligned}
$

Summary

The EMF of a cell is the driving force that enables the flow of electric current in a circuit by converting chemical energy into electrical energy. The potential difference, which is the work done by a unit charge, depends on the current and resistance within the circuit. Understanding these concepts is key to solving problems related to electric circuits, such as determining current flow, potential differences, and the behaviour of batteries and resistors in various configurations.