3.11 Cells, emf, Internal Resistance

Cell 

A cell converts chemical energy into electrical energy. 

Electrolytic Cell

A simple device maintain a steady current in an electric circuit is the electrolytic cell. Basically a cell has two electrodes, called the positive (P) and the negative (N), as shown in Fig. 3.18. They are immersed in an electrolytic solution. Dipped in the solution, the electrodes exchange charges with the electrolyte. The positive electrode has a potential difference \(V_{+}\left(V_{+}>0\right)\) between itself and the electrolyte solution immediately adjacent to it marked A in the figure. Similarly, the negative electrode develops a negative potential \(-\left(V_{-}\right)\left(V_{-} \geq 0\right)\) relative to the electrolyte adjacent to it, marked as \(B\) in the figure. When there is no current, the electrolyte has the same potential throughout, so that the potential difference between P and N is \(V_{+}-\left(-V_{-}\right)=V_{+}+V_{-}\). This difference is called the electromotive force (emf) of the cell and is denoted by \(\varepsilon\). Thus
\(
\varepsilon=V_{+}+V_{-}>0
\)

Note that \(\varepsilon\) is, actually, a potential difference and not a force. The name emf, however, is used because of historical reasons and was given at a time when the phenomenon was not understood properly.

To understand the significance of \(\varepsilon\), consider a resistor \(R\) connected across the cell (Fig. 3.18). A current \(I\) flows across \(R\) from C to D . As explained before, a steady current is maintained because current flows from N to P through the electrolyte. Clearly, across the electrolyte the same current flows through the electrolyte but from N to P, whereas through \(R\), it flows from P to N.

Representation of Battery

Ideal Cell

Cell in which there is no heating effect (internal resistance is 0). 

Non-Ideal Cell

Cell in which there is a heating effect inside due to opposition to the current flow internally (there is internal resistance). The electrolyte through which a current flows has a finite resistance \(r\), called the internal resistance. 

Potential difference \((V)\) across the terminals of a cell

Case-I: Battery Acting as a Source (or Battery is Discharging)
\(
\begin{aligned}
& V _{ A }- V _{ B }=\varepsilon- ir \\
& V=V_{AB}=V _{ A }- V _{ B } \text { it is also called terminal voltage}.
\end{aligned}
\)
The rate at which the chemical energy of the cell is consumed \(=\varepsilon i\)
The rate at which heat is generated inside the battery or cell \(=i^2 r\)
\(
\begin{aligned}
\text { electric power output } & =\varepsilon i-i^2 r \\
& =(\varepsilon- ir ) i
\end{aligned}
\)

Case II : Battery Acting as a Load (or Battery Charging)
\(
V=V_{AB}=V _{ A }- V _{ B }=\varepsilon+\text { ir }
\)

the rate at which chemical energy stored in the cell \(=\varepsilon i\)
thermal power inside the cell \(\quad=i^2 r\)
electric power input \(=\varepsilon i+i^2 r =(\varepsilon+ ir ) i =\left( V _{ A }- V _{ B }\right) i\)

Case III : When cell is in open circuit
\(i =0\) as resistance of open circuit is infinite \((\infty)\).
So \(V =\varepsilon\), so the open circuit terminal voltage difference is equal to emf of the cell.

Case IV : Short-circuiting

Two points in an electric circuit directly connected by a conducting wire are called short-circuited, under such condition both points are at the same potential.
When cell is short-circuited
\(i =\frac{\varepsilon}{ r }\) and \(V =0\), short circuit current of a cell is maximum. The potential at all points of a wire of zero resistance will be the same.

Earthing :

If some point of the circuit is earthed then its potential is assumed to be zero. It is represented by the symbol shown below.

Electromotive force (E. M. F.)
The potential difference across the terminals of a cell when it is not delivering any current is called the emf of the cell. The energy given by the cell in the flow of unit charge in the whole circuit (including the cell) is called the emf of the cell.

Definition:  It is the work done by the battery for the flow of 1-coulomb charge from the lower potential terminal to the higher potential terminal inside the battery. Alternately we can say the electromotive force of a cell is equal to the potential difference between its terminals when no current is passing through the circuit.

emf depends on :

• nature of electrolyte
• metal of electrodes

emf does not depend on :

• area of plates
• distance between the electrodes
• quantity of electrolyte
• size of cell

Example 47: The reading on a high resistance voltmeter when a cell is connected across it is 3 V . When the terminals of the cell are also connected to a resistance of \(4 \Omega\), then the voltmeter reading drops to 1.2 V. Find the internal resistance of the cell.

Solution: Given, \(E=3 V, R=4 \Omega\) and \(V=12 V\)
As, internal resistance,
\(
r=R\left(\frac{E-V}{V}\right)=4\left(\frac{3-1.2}{1.2}\right) \Omega=6 \Omega
\)

Example 48: A battery of emf 2 V and internal resistance \(r\) is connected in series with a resistor of \(10 \Omega\) through an ammeter of resistance \(2 \Omega\). The ammeter reads 50 mA. Draw the circuit diagram and calculate the value of \(r\).

Solution:

\(
\text { Total resistance }=10+2+r=(12+r) \Omega
\)

Now, current \(=50 mA=50 \times 10^{-3} A\) and \(emf =2 V\).
\(
\begin{array}{rlrl}
\text { So, } & \text { resistance } =\frac{\text { emf }}{\text { current }} \\
12+r & =\frac{2}{50 \times 10^{-3}}=40 \\
\Rightarrow & r =40-12=28 \Omega
\end{array}
\)

Example 49: A voltmeter of resistance \(994 \Omega\) is connected across a cell of emf \(1 V\) and internal resistance \(6 \Omega\). Find the potential difference across the voltmeter, that across the terminals of the cell and percentage error in the reading of the voltmeter.

Solution: Given, \(E=1 V, r=6 \Omega\)
and resistance of voltmeter, \(R=994 \Omega\)

The current in the circuit is
\(
I=\frac{E}{R+r}=\frac{1}{(994+6)}=1 \times 10^{-3} A
\)
The potential difference across the voltmeter is
\(
\begin{aligned}
V & =I R=1 \times 10^{-3} \times 994 \\
& =9.94 \times 10^{-1} V
\end{aligned}
\)
The same will be the potential difference across the terminals of the cell. The voltmeter used to measure the emf of the cell will read 0.994 V . Hence, the percentage error is
\(
\frac{E-V}{E} \times 100=\frac{1-0.994}{1} \times 100=0.6 \%
\)

Example 50: Find the current drawn from a cell of emf 2 V and internal resistance \(2 \Omega\) connected to the network given below.

Solution: The equivalent circuit is shown below

Resistance in \(\operatorname{arm} A B=1 \Omega\)
Resistance in \(\operatorname{arm} P Q=\frac{1 \times 1}{1+1}+\frac{1 \times 1}{1+1}=\frac{1}{2}+\frac{1}{2}=1 \Omega\)
Resistance in \(\operatorname{arm} D C=1 \Omega\)
These three resistances are connected in parallel.
Their equivalent resistance \(R\) is given by
\(
\begin{aligned}
& \frac{1}{R}=\frac{1}{1}+\frac{1}{1}+\frac{1}{1}=\frac{3}{1} \\
& R=\frac{1}{3} \Omega
\end{aligned}
\)
Current drawn from the cell,
\(
\begin{aligned}
& I=\frac{E}{R+r}=\frac{2}{\left(\frac{1}{3}+2\right)} \\
& =\frac{2 \times 3}{7}=\frac{6}{7} A
\end{aligned}
\)

Example 51: In the given electric circuit find
(a) current
(b) power output
(c) relation between r and R so that the electric power output (that means power given to \(R\) ) is maximum.
(d) value of maximum power output.
(e) plot graph between power and resistance of the load
(f) From the graph we see that for a given power output there exist two values of external resistance, prove that the product of these resistances equals \(r ^2\).
(g) what is the efficiency of the cell when it is used to supply maximum power?

Solution: (a) In the circuit shown if we assume that potential at \(A\) is zero then the potential at \(B\) is
\(\varepsilon-ir\). Now since the connecting wires are of zero resistance
\(
\therefore \quad V _{ D }= V _{ A }=0 \quad \Rightarrow \quad V _{ C }= V _{ B }=\varepsilon- ir
\)
Now current through CD is also \(i\)
\((\because\) it’s in series with the cell).
\(
\therefore \quad i =\frac{V_C-V_D}{R}=\frac{(\varepsilon-i r)-0}{R}
\)
Current \(i =\frac{\varepsilon}{r+R}\)
(b) Power output \(P =i^2 R =\frac{\varepsilon^2}{(r+R)^2} \cdot R\)
(c) To calculate maximum power we know at maxima the slope is zero. Therefore differentiating the power equation with respect to the load we get,
\(
\frac{ dP }{ dR }=\frac{\varepsilon^2}{(r+R)^2}-\frac{2 \varepsilon^2 R}{(r+R)^3}=\frac{\varepsilon^2}{(R+r)^3}[R+r-2 R]
\)
for maximum power transer to load \(R\)
\(
\frac{d P}{d R}=0 \quad \Rightarrow r + R -2 R =0 \quad \Rightarrow r = R
\)
Note: Here for maximum power output load resistance should be equal to internal resistance of the source (in this case battery).
(d) \(P_{\max }=\frac{\varepsilon^2}{4 r}\)
(e) Graph between ‘P’ and \(R\) maximum power output at \(R = r\)
\(
P _{\max }=\frac{\varepsilon^2}{4 r } \Rightarrow i=\frac{\varepsilon}{r+R}
\)

(f) Power output
\(
\begin{aligned}
& P =\frac{\varepsilon^2 R}{(r+R)^2} \\
& P \left( r ^2+2 rR + R ^2\right)=\varepsilon^2 R \\
& R ^2+\left(2 r -\frac{\varepsilon^2}{P}\right) R + r ^2=0
\end{aligned}
\)
above quadratic equation in R has two roots \(R _1\) and \(R _2\) for given values of \(\varepsilon, P\) and \(r\) such that \(\therefore \quad R_1 R_2=r^2 \quad\) (product of roots) \(\Rightarrow r^2=R_1 R_2\)
(g) Power of battery spent
\(
=\frac{\varepsilon^2}{(r+r)^2} \cdot 2 r =\frac{\varepsilon^2}{2 r}
\)
power (output)
\(
=\left(\frac{\varepsilon}{r+r}\right)^2 \times r=\frac{\varepsilon^2}{4 r}
\)
\(
\text { Efficiency }=\frac{\text { power output }}{\text { total power spent by cell }}=\frac{\frac{\varepsilon^2}{4 r} \times 100}{\frac{\varepsilon^2}{2 r}}=\frac{1}{2} \times 100=50 \%
\)

Example 52: In the figure given beside find out the current in the wire BD

Solution: Let at point \(D\) potential \(=0\) and write the potential of other points then
current in wire \(AD =\frac{10}{2}=5 A\) from A to D
current in wire \(CB =\frac{20}{5}=4 A\) from C to B
\(\therefore\) current in wire \(BD =1 A\) from D to B

Example 53: A network of resistors is connected to a 16 V battery with an internal resistance of \(1 \Omega\), as shown in Fig. 3.19: (a) Compute the equivalent resistance of the network. (b) Obtain the current in each resistor. (c) Obtain the voltage drops \(V_{A B}, V_{B C}\) and \(V_{C D}\).

Solution: (a) The network is a simple series and parallel combination of resistors. First the two \(4 \Omega\) resistors in parallel are equivalent to a resistor \(=[(4 \times 4) /(4+4)] \Omega=2 \Omega\)
In the same way, the \(12 \Omega\) and \(6 \Omega\) resistors in parallel are equivalent to a resistor of
\(
[(12 \times 6) /(12+6)] \Omega=4 \Omega
\)
The equivalent resistance \(R\) of the network is obtained by combining these resistors ( \(2 \Omega\) and \(4 \Omega\) ) with \(1 \Omega\) in series, that is.
\(
R=2 \Omega+4 \Omega+1 \Omega=7 \Omega
\)
(b) The total current \(I\) in the circuit is
\(
I=\frac{\varepsilon}{R+r}=\frac{16 V}{(7+1) \Omega}=2 A
\)
Consider the resistors between A and B. If \(I_1\) is the current in one of the \(4 \Omega\) resistors and \(I_2\) the current in the other,
\(
I_1 \times 4=I_2 \times 4
\)
that is, \(I_1=I_2\), which is otherwise obvious from the symmetry of the two arms. But \(I_1+I_2=I=2 A\). Thus,
\(
I_1=I_2=1 A
\)
that is, current in each \(4 \Omega\) resistor is 1 A. Current in \(1 \Omega\) resistor between B and C would be 2 A.
Now, consider the resistances between C and D . If \(I_3\) is the current in the \(12 \Omega\) resistor, and \(I_4\) in the \(6 \Omega\) resistor,
\(
I_3 \times 12=I_4 \times 6 \text {, 1.e., } I_4=2 I_3
\)
But, \(I_3+I_4=I=2 A\)
Thus, \(I_3=\left(\frac{2}{3}\right) A , I_4=\left(\frac{4}{3}\right) A\)
that is, the current in the \(12 \Omega\) resistor is \((2 / 3) A\), while the current In the \(6 \Omega\) resistor is \((4 / 3) A\).
(c) The voltage drop across AB is
\(
V_{A B}=I_1 \times 4=1 A \times 4 \Omega=4 V,
\)
This can also be obtained by multiplying the total current between A and B by the equivalent resistance between A and B , that is,
\(
V_{A B}=2 A \times 2 \Omega=4 V
\)
The voltage drop across BC is
\(
V_{B C}=2 A \times 1 \Omega=2 V
\)
Finally, the voltage drop across \(C D\) is
\(
V_{C D}=12 \Omega \times I_3=12 \Omega \times\left(\frac{2}{3}\right) A =8 V \text {. }
\)
This can alternately be obtained by multiplying the total current between C and D by the equivalent resistance between C and D, that is,
\(
V_{C D}=2 A \times 4 \Omega=8 V
\)
Note that the total voltage drop across AD is \(4 V+2 V+8 V=14 V\). Thus, the terminal voltage of the battery is 14 V, while its emf is 16 V . The loss of the voltage \((=2 V)\) is accounted for by the internal resistance \(1 \Omega\) of the battery [ \(2 A \times 1 \Omega=2 V]\).