5.1 Combination of cells

1.1 Current in different situations

2.1 Expression for drift velocity
2.2 Mobility $\left( \mu \right)$

3.1 Temperature dependence of resistance
3.2 Resistivities of different materials
3.3 Limitations of ohm's law

5.1 Combination of cells

6.1 Maximum power transfer theorem
6.2 Kirchhoff's law

9.1 Comparison of emfs of two primary cells.
9.2 Determination of Internal resistance of a cell using potentiometer

10.1 Ammeter
10.2 Voltmeter

If $n$ cells of emf $\left( {{\xi _1},{\xi _2}.....{\xi _n}} \right)$ and internal resistance $\left( {{r_1},{r_2}.....{r_n}} \right)$ are connected in series to external resistance $R$.

Then Equivalent emf, $({\xi _{eq}})$, $$ \Rightarrow {\xi _{eq}} = {\xi _1} + {\xi _2}..... + {\xi _n}$$

Also, equivalent internal resistance $({r_{eq}})$ ,$$ \Rightarrow {r_{eq}} = {r_1} + {r_2} + ..... + {r_n}$$

Total resistance $(R_{eq})$ of the circuit, $$R_{eq}=R + r_{eq}$$

The above circuit becomes as shown.

The current $(I)$ in the circuit, $$I = \frac{{\xi_{eq} }}{{R + r_{eq}}}$$

Equivalent emf $({\xi _{eq}})$, $$ \Rightarrow {\xi _{eq}} = n\xi $$

Equivalent internal resistance $(r_{eq})$ ,$${r_{eq}} =nr$$

The current $(I)$ in the circuit, $$I = \frac{n{\xi}}{{R + nr}}$$

(i). If $R \gg nr$, then $I = \frac{{n\xi }}{R}$= n times the current due to single cell

(ii).If $R \ll nr$, then $I = \frac{{n\xi }}{{nr}} = \frac{\xi }{r}$ = current given by single cell.

Also,

When the polarity of one cell having emf ${\xi _2}$ and internal resistance ${{r_2}}$ is reversed as shown in the figure.

Then, Equivalent emf $({\xi _{eq}})$, $${\xi _{eq}} = {\xi _1} - {\xi _2}..... + {\xi _n}$$

Also, equivalent internal resistance $({r_{eq}})$ ,$${r_{eq}} = {r_1} + {r_2} + ..... + {r_n}$$

$n$ identical cells connected in parallel.

Let there $n$ identical rows each containing a cell of emf $\xi$ and internal resistane $r$ connected with an external resistance $R$ as shown in the figure.

Let ${I_1},{I_2}...{I_n}$ be the current in each row as shown in the figure.

Applying Kirchhoff's junction law at point $A$ we get, $$ \Rightarrow {I_1} + {I_2} + ... + {I_n} = I$$

Applying Kirchhoff's loop law for each loop we get, $$\begin{equation} \begin{aligned} \xi - {I_1}r - IR = 0 \\ \xi - {I_2}r - IR = 0 \\ ..... \\ ..... \\ \xi - {I_n}r - IR = 0 \\\end{aligned} \end{equation} $$

Adding the above equations we get, $$\begin{equation} \begin{aligned} \Rightarrow n\xi - \left( {{I_1} + {I_2} + ... + {I_n}} \right)r - nIR = 0 \\ \Rightarrow n\xi - Ir - nIR = 0 \\ \Rightarrow I = \frac{{n\xi }}{{nR + r}} \\\end{aligned} \end{equation} $$

(i) If $R \ll \frac{r}{n}$ , then $I = \frac{{n\xi }}{r}$ = $n$ times current due to single cell.

(ii) If $R \gg \frac{r}{n}$, then $I = \frac{\xi }{R}$, current given by single cell.

Let there $m$ identical rows each containing $n$ cells each of emf $\xi$ and internal resistane $r$ connected with an external resistance $R$ as shown in the figure.

Let ${I_1},{I_2}...{I_n}$ be the current in each row as shown in the figure.

Applying Kirchhoff's junction law at point $A$ we get, $$ \Rightarrow {I_1} + {I_2} + ... + {I_m} = I$$

Applying Kirchhoff's loop law for each loop we get, $$\begin{equation} \begin{aligned} n\xi - {I_1}(nr) - IR = 0 \\ n\xi - {I_2}(nr) - IR = 0 \\ ..... \\ ..... \\ n\xi - {I_m}(nr) - IR = 0 \\\end{aligned} \end{equation} $$

Adding the above equations we get, $$\begin{equation} \begin{aligned} \Rightarrow mn\xi - \left( {{I_1} + {I_2} + ... + {I_m}} \right)(nr)mIR = 0 \\ \Rightarrow mn\xi - Inr - mIR = 0 \\ \Rightarrow I = \frac{{mn\xi }}{{mR + nr}} \\\end{aligned} \end{equation} $$

Question 10. A wire of resistance $4R$ is bent in the form of circle. What is the effective resistance between ends of diameter?

Solution: When the wire of resistance $4R$ is bend in the form of a circle, then the current will flow as shown in the figure.

So, the above situation can be understood as, parallel combination of resistance as showed in the diagram below.

The resistance are in parallel with each other. So the resistance between end $A$ and $B$ is, $$\begin{equation} \begin{aligned} \Rightarrow \frac{1}{{{R_{eq}}}} = \frac{1}{{2R}} + \frac{1}{{2R}} \\ \Rightarrow \frac{1}{{{R_{eq}}}} = \frac{1}{R} \\ \Rightarrow {R_{eq}} = R \\\end{aligned} \end{equation} $$

Question 11. In the given network of resistors. Find the equivalent resistance between point $A$ and $B$.

Resistors $AD$ and $DC$ are in series. So, $$\begin{equation} \begin{aligned} \Rightarrow {R_{ADC}} = {R_{AD}} + {R_{DC}} \\ \Rightarrow {R_{ADC}} = (3 + 7)\Omega \\ \Rightarrow {R_{ADC}} = 10\Omega \\\end{aligned} \end{equation} $$

Resistors $R_{ADC}$ and $R_{AC}$ are in parallel. So, $$\begin{equation} \begin{aligned} \Rightarrow \frac{1}{R} = \frac{1}{{{R_{ADC}}}} + \frac{1}{{{R_{AC}}}} \\ \Rightarrow \frac{1}{R} = \frac{1}{{10}} + \frac{1}{{10}} = \frac{2}{{10}} \\ \Rightarrow R = 5\Omega \\\end{aligned} \end{equation} $$

Now, resistors $R$ and $R_{CB}$ are in series. So, $$\begin{equation} \begin{aligned} \Rightarrow R' = R + {R_{CB}} \\ \Rightarrow R' = (5 + 5)\Omega \\ \Rightarrow R' = 10\Omega \\\end{aligned} \end{equation} $$

Finally resistors $R'$ and $R_{AB}$ are in parallel. So, $$\begin{equation} \begin{aligned} \Rightarrow \frac{1}{R_{AB}} = \frac{1}{R'} + \frac{1}{{{R_{AB}}}} \\ \Rightarrow \frac{1}{R_{AB}} = \frac{1}{{10}} + \frac{1}{{10}} = \frac{2}{{10}} \\ \Rightarrow {R_{AB}} = 5\Omega \\\end{aligned} \end{equation} $$

Question 12. Calculate total current flowing through the circuit.

Solution. The circuit diagram can be redrawn as shown in figure.

Now, the equivalent resistance can be very calculated. $$R_{eq}=2\Omega$$

Thus, Current flowing through circuit is $$I = \frac{V}{R} = \frac{{10V}}{{2\Omega }} = 5A$$

Question 13. In the circuit shown, calculate the effective resistance.

Solution. Effective resistance between $E$ and $F$ ,

Resistance ${2\Omega }$, ${4\Omega }$ and ${2\Omega }$ are in series and their equivalent is in parallel with ${8\Omega }$

Effective resistance is $$R_{EF} = \frac{{8 \times 8}}{{8 + 8}} = 4\Omega $$

Similarly, effective resistance between $C$ and $D$ is $$R_{CD} = \frac{{8 \times 8}}{{8 + 8}} = 4\Omega $$

Then finally resistance ${3\Omega }$, ${4\Omega }$ and ${2\Omega }$ are in series.

Equivalent resistance of the circuit $$R_{AB} = 3 + 4 + 2 = 9\Omega $$

Question 14. A cell of emf $4V$ and internal resistance 0.2$\Omega $ is connected to a 7.8$\Omega $ external resistance. What will be the potential difference across the terminals of cell?

Given $\xi = 4V$ , $r =0.2\Omega $ , $R=7.8\Omega $

Current, $$I = \frac{\xi }{{R + r}} = \frac{4}{{7.8 + 0.2}} = 0.5A$$

Potential difference across terminals $AB$ of cell, $$V_{AB} = IR = 0.5 \times 7.8 = 3.9V$$

or

Potential difference across terminals $AB$ of cell, $$V_{AB} = \xi - Ir = 4-0.2 \times 0.5 = 3.9V$$