We have learn now what a resistance it. Now let’s see how the combination of resistances can be used in a circuit to control the flow of current and voltage division. Majorly the combination of these resistances are of 2 types: Series and Parallel. Here we will be taken both in detail below.

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## Combination of Resistances

A series circuit comprises a path along which the whole current flows through each component. A parallel circuit comprises branches so that the current divides and only part of it flows through any branch. The voltage, or potential difference, across each branch of a parallel circuit is the same, but the currents may vary. In a home electrical circuit, for instance, the same voltage is applied across each light or appliance, but each of these loads draws a different amount of current, according to its power requirements. A number of similar batteries connected in parallel provides greater current than a single battery, but the voltage is the same as for a single battery. [source]

Let’s start with Combination of resistances with series combination at first then parallel combination.

### Series Combination

Resistances are said to be connected in series between two points if they provide only a single path between two points. Resistances are connected in series if same current flows through each resistance when some potential difference is applied across the combination.

Potential difference across each resistance is different and is directly proportional to its resistance ${\rm{V}} \propto {\rm{R}}.$

So $\quad {{\rm{V}}_1} = {\rm{I}}{{\rm{R}}_1};\quad {{\rm{V}}_2} = {\rm{I}}{R_2}\quad $ and $\quad {{\rm{V}}_3} = {\rm{I}}{{\rm{R}}_3}$

- The series combination obeys law of conservation of energy

So $V = {V_1} + {V_2} + {V_3} = I\left( {{R_1} + {R_2} + {R_3}} \right)$

Equivalent resistance ${{\rm{R}}_{\rm{s}}} = {{\rm{V}} \over {\rm{I}}} = {{\rm{R}}_1} + {{\rm{R}}_2} + {{\rm{R}}_3}$ - The equivalent resistance is equal to sum of individual resistances.
- The equivalent resistance is greater than largest of individual resistance.
- The resistances are connected in series
- to increase the resistance and
- to divide large potential difference across many resistances.

- In ‘n’ identical resistances R are connected in series then the equivalent resistance R
_{s}= nR - This combination is used in resistance boxes and sometimes in decorative bulbs.
- In resistances connected in series if one resistance get open the current in whole circuit will become zero.

### Parallel Combination

Resistances are said to be connected in parallel between two points if it is possible to proceed from one point to another along different paths.

Resistances are said to be in parallel if potential across each resistance is same and equal to applied potential.

Current through each resistance is different and is inversely proportional to resistance of resistor.

$l \propto 1/R.$ So $\quad {I_1} = {V \over {{R_1}}},{I_2} = {V \over {{R_2}}}\quad $ and $\quad {I_3} = {V \over {{R_3}}}$

- The parallel combination obeys the conservation of charge.

So $\quad {\rm{I}} = {{\rm{I}}_1} + {{\rm{I}}_2} + {{\rm{I}}_3} = {\rm{V}}\left( {{1 \over {{{\rm{R}}_1}}} + {1 \over {{{\rm{R}}_2}}} + {1 \over {{{\rm{R}}_3}}}} \right)$

Reciprocal of equivalent resistance ${1 \over {{{\rm{R}}_{\rm{p}}}}} = {{\rm{I}} \over {\rm{V}}} = {1 \over {{{\rm{R}}_1}}} + {1 \over {{{\rm{R}}_2}}} + {1 \over {{{\rm{R}}_3}}}$ - The reciprocal of equivalent resistance is equal to sum of reciprocal of individual resistances.
- The equivalent resistance is smaller than smallest of individual resistance.
- The resistances are connected in parallel to decrease resistance.
- If ‘n’ identical resistances $R$ are connected in parallel then equivalent resistance ${R_p} = R/n$
- This combination is used in household electrical appliances.
- In resistances connected in parallel if one resistance becomes open then also all others will work as usual.
- In case of two resistances in parallel ${{{{\rm{I}}_1}} \over {{{\rm{I}}_2}}} = {{{{\rm{R}}_2}} \over {{{\rm{R}}_1}}}\quad $ and $\quad {{\rm{I}}_1} + {{\rm{I}}_2} = {\rm{I}}$

So $\quad {{\rm{I}}_1} = {{{{\rm{R}}_2}{\rm{I}}} \over {{{\rm{R}}_1} + {{\rm{R}}_2}}}\quad $ and $\quad {{\rm{I}}_2} = {{{{\rm{R}}_1}{\rm{I}}} \over {{{\rm{R}}_1} + {{\rm{R}}_2}}}$

## Solved Examples on Combination of Resistances

**Sol.**Point 1 and 3 are at same potential. Similarly point 2 and 4 are at same potential. Joining resistance between 1 and 2,2 and 3,3 and 4 we find that they all are in parallel.

So equivalent resistance ${1 \over {{R_p}}} = {1 \over {2R}} + {1 \over {2R}} + {1 \over R} = {2 \over R}\quad $ or $\quad {R_p} = R/2$

**Sol.**Resistance of arm ${\rm{ADC}} = 3 + 7 = 10\Omega $ .

This is in parallel to 10$\Omega $ of arm AC. Their parallel combination gives equivalent resistance of 5$\Omega$

So effective resistance between ${\rm{B}}$ and ${\rm{C}}$

${{\rm{R}}_{{\rm{eq}}}} = {{14 \times 5} \over {14 + 5}} = {{70} \over {19}} = 3.684\Omega $

**Sol.**Let the two resistances be ${R_1}$ and ${R_2}$

When connected in series ${R_1} + {R_2} = 40$

When connected in parallel ${{{R_1}{R_2}} \over {{R_1} + {R_2}}} = 7.5\quad $ so $\quad {R_1}{R_2} = 7.5 \times 40 = 300$

${\left( {{R_1} – {R_2}} \right)^2} = {\left( {{R_1} + {R_2}} \right)^2} – 4{R_1}{R_2} = {40^2} – 4 \times 300 = 400$

So $\quad {R_1} – {R_2} = 20\Omega $

Solving 1 and 2 we have ${R_1} = 30\Omega $ and ${R_2} = 10\Omega $

**Sol.**By symmetry potential at point 4 and 5 is same. Similarly potential at point 3 and 6 is same. The equivalent circuits can be drawn

as :

The equivalent resistance between a and b is ${{\rm{R}}_{{\rm{eq}}}} = {{{\rm{R}} \times {7 \over 5}{\rm{R}}} \over {{\rm{R}} + {7 \over 5}{\rm{R}}}} = {7 \over {12}}{\rm{R}}$

**Sol.**The equivalent circuits can be drawn as

The equivalent resistance of circuit is ${{\rm{R}}_{{\rm{eq}}}} = {{10} \over 3}\Omega $

The current ${\rm{I}} = {{\rm{E}} \over {{{\rm{R}}_{{\rm{eq}}}}}} = {{12} \over {10/3}} = 3.6{\rm{A}}$

$(1)\,4\Omega $

$(2)\,8\Omega $

$(3)\,6\Omega $

$(4)\,2\Omega $

**Sol.**$\quad {\rm{R}} = {{\left( {{{\rm{R}}_1} + {{\rm{R}}_2}} \right)} \over 2} + {1 \over 2}{\left[ {{{\left( {{{\rm{R}}_1} + {{\rm{R}}_2}} \right)}^2} + 4{{\rm{R}}_3}\left( {{{\rm{R}}_1} + {{\rm{R}}_2}} \right)} \right]^{{1 \over 2}}} \ldots \ldots \ldots $ (1)

${\rm{R}}1 = 1\Omega ,{{\rm{R}}_2} = 0.{{\rm{R}}_3} = 2\Omega $

From eqs. $(1)$ and $(2)$

${\rm{R}} = {1 \over 2} + {1 \over 2}\left[ {1 + 4 \times 2 \times 4{1 \over 2}[1 + 3] = 2\Omega } \right.$

**Sol.**After distribution of current using loop rule for

$\quad $ Loop a $ – \left( {{\rm{I}} – {{\rm{I}}_1}} \right){\rm{R}} – {{\rm{I}}_2}{\rm{R}} + {{\rm{I}}_1}{\rm{R}} = 0\quad $ or $\quad 2{{\rm{I}}_1} – {{\rm{I}}_2} = {\rm{I}}$

Loop b $ – \left( {1 – {{\rm{I}}_1} – {{\rm{I}}_2}} \right){\rm{R}} + {{\rm{I}}_3}{\rm{R}} + {{\rm{I}}_2}{\rm{R}} = 0\quad $ or $\quad {{\rm{I}}_1} + 2{{\rm{I}}_2} + {{\rm{I}}_3} = {\rm{I}}$

Loop ${\rm{c}} – {{\rm{I}}_3}{\rm{R}} – \left( {{\rm{I}} – {{\rm{I}}_1} – {{\rm{I}}_2} + {{\rm{I}}_3}} \right){\rm{R}} + \left( {{{\rm{I}}_1} + {{\rm{I}}_2} – {{\rm{I}}_3}} \right){\rm{R}} = 0\quad $ or $\quad 2{{\rm{I}}_1} + 2{{\rm{I}}_2} – 3{{\rm{I}}_3} = {\rm{I}}$

Solving for ${{\rm{I}}_1},{{\rm{I}}_2}$ and ${{\rm{I}}_3}$ we get,

${{\rm{I}}_1} = {4 \over 7}{\rm{I}},{{\rm{I}}_2} = {{\rm{I}} \over 7},{{\rm{I}}_3} = {{\rm{I}} \over 7}$

Here ${{\rm{I}}_2} = {{\rm{I}}_3}$ so, no distribution of current takes place at D so it is a pseudo junction.

The equivalent circuit becomes.

Resistance of triangle DCE is R parallel to $({\rm{R}} + {\rm{R}}) = {2 \over 3}{\rm{R}}$

Resistance of network other than ${\rm{AB}} = {\rm{R}} + {2 \over 3}{\rm{R}} + {\rm{R}} = {8 \over 3}{\rm{R}}$

Equivalent resistance between ${\rm{AB}}$

${{\rm{R}}_{{\rm{eq}}}} = {{2{\rm{R}} \times {{8{\rm{R}}} \over 3}} \over {2{\rm{R}} + {{8{\rm{R}}} \over 3}}} = {8 \over 7}{\rm{R}}$

**Sol.**Ratio of upper resistances 5 : 10 : 15=1 : 2 : 3

Ratio of lower resistances 10 : 20 : 30=1 : 2 : 3

The ratio is same so resistance in middle are not useful.

Equivalent resistance =(5+10+15)| |(10+20+30)

So ${R_{{\rm{ eq }}}} = {{30 \times 60} \over {30 + 60}} = 20\Omega $

**Sol.**Using voltage rule from ${\rm{b}}$ to ${\rm{d}}$ via c we get

${{\rm{V}}_{\rm{b}}} – 1 \times 2 + 4 \times 1 = {{\rm{V}}_{\rm{d}}}\quad $ or $\quad {{\rm{V}}_{\rm{d}}} – {{\rm{V}}_{\rm{b}}} = 2$ volt.