Parallel combination of resistances: If a number of resistances are connected in between two common points so that each of them provides a separate path for current, then they are said to be connected in parallel.
As shown in Fig, consider three resistances R₁,R₂ and R₃ connected in parallel. Suppose a current I flows through the circuit when a cell of voltage V is connected across the combination. The current I at point A is divided into three parts I₁,I₂ and I₃ through the resistances R₁, R₂ and R₃ respectively. These three parts recombine at point B to give the same current I.

Fig. Resistances connected in parallel
∴    I = I₁ + I₂ + I₃
As all the three resistances have been connected between the same two points A and B, so voltage V across each of them is same. By Ohm’s law,

If R_p be the equivalent resistance of the parallel combination, then,
                    
But             
      
or                

Laws of resistances in parallel:
(i)    Voltage across each resistance is same and is equal to the applied voltage.
(ii)    Total current = Sum of the currents through the individual resistances.
(iii)    Currents through various resistances are inversely proportional to the individual resistances.
(iv)    Reciprocal of equivalent resistance = Sum of reciprocals of individual resistances.
(v)    Equivalent resistance is less than the smallest individual resistance.

Series combination of resistances: If a number of resistances are joined end to end so that the same current flows through each of them in succession, then the resistances are said to be connected in series.

Fig. Resistances connected in series
As shown in Fig, consider three resistances R₁, R₂ and R₃ connected in series. Suppose a current I flows through the circuit when a cell of voltage V is connected across the combination.
By Ohm’s law, the potential differences across the three resistances will be,

or

Laws of resistances in series:
(i)    Current through each resistance is same.
(ii)    Total voltage across the combination = Sum of the voltage drops.
(iii)    Voltage drop across any resistor is proportional to its resistance.
(iv)    Equivalent resistance = Sum of the individual resistances.
(v)    Equivalent resistance is larger than the largest individual resistance.

Given, two resistances are connected in parallel to give an equivalent resistance of 2 Ω. The same resistors when connected in series gives an equivalent resistance of 9 Ω. 

Let, R₁ and R₂ be two resistances.
 
So, we have 
              R₁ + R₂ = 9 Ω  and 

              

$\Rightarrow$            $\frac{{\mathrm{R}}_1+{\mathrm{R}}_2}{{\mathrm{R}}_1{\mathrm{R}}_2} = \frac{1}{2}$

$\Rightarrow$               $\frac{9}{{\mathrm{R}}_1{\mathrm{R}}_2} = \frac{1}{2}$ 

$\Rightarrow$                 R₁R₂ = 18 

Now, using the mathematical identity of (a-b)², we have 

(R₁ – R₂)² = (R₁ + R₂)² – 4R₁R₂ 
                = (9)² – 4 x 18
                = 81 – 72
                = 9 

∴       R₁ – R₂ = 3    ...(1) 

         R₁ + R₂= 9    ...(2) 

Adding (1) and (2), we get 

                  2R₁ = 12

$\Rightarrow$                R₁ = 6 Ω. 

and               R₂ = 9 – 6 = 3 Ω. 

The resistances given are: 

R₁ = 5 Ω, R₂ = 10 Ω, R₃ = 30 Ω 

Battery voltage gives us the resistance across each resistor i.e., = 12 V 

(a) Now, using Ohm’s law, 

Current through ${\mathrm{R}}_1, {\mathrm{I}}_1 = \frac{\mathrm{V}}{{\mathrm{R}}_1} = \frac{12}{5} = 2.4 \mathrm{A}.$ 

Current through ${\mathrm{R}}_2, {\mathrm{I}}_2 = \frac{\mathrm{V}}{{\mathrm{R}}_2} = \frac{12}{10} = 1.2 \mathrm{A}.$ 

Current through ${\mathrm{R}}_3, {\mathrm{I}}_3 = \frac{\mathrm{V}}{{\mathrm{R}}_3} = \frac{12}{30} = 0.4 \mathrm{A}.$

(b) Total current across the circuit is the sum of all the currents flowing through the individual resistors.

I = I₁ + I₂ + I₃ = 2.4 + 1.2 + 0.4 = 4 A. 

(c) Equivalent resistance across the parallel combination is,

                      $$\begin{gathered} \frac{1}{\mathrm{R}}= \frac{1}{{\mathrm{R}}_1}+\frac{1}{{\mathrm{R}}_2}+\frac{1}{{\mathrm{R}}_3} \\ = \frac{1}{5}+\frac{1}{10}+\frac{1}{30} \\ = \frac{1}{3} \end{gathered}$$ 

i.e.,                 $\mathrm{R} = 3 \mathrm{\Omega }.$

For the circuit given, we have

(a)  Current through $5 \mathrm{\Omega }$ resistor, ${\mathrm{I}}_1 = \frac{\mathrm{V}}{{\mathrm{R}}_1} = \frac{6 \mathrm{V}}{5 \mathrm{\Omega }} = 1.2 \mathrm{A}$
       Current through $10 \mathrm{\Omega }$ resistor, ${\mathrm{I}}_2 = \frac{\mathrm{V}}{{\mathrm{R}}_2} = \frac{6 \mathrm{V}}{10 \mathrm{\Omega }} =0.6 \mathrm{A}$
       Current through $30 \mathrm{\Omega }$ resistor, ${\mathrm{I}}_3$$= \frac{\mathrm{V}}{{\mathrm{R}}_3} = \frac{6 \mathrm{V}}{30 \mathrm{\Omega }} = 0.2 \mathrm{A}.$ 

(b) Total current in the circuit is given by, 

I = I₁ + I₂ + I₃ = 1.2 + 0.6 + 0.2 = 2.0 A. 

(c) Total effective resistance R is given by, 

         $$\begin{gathered} \frac{1}{\mathrm{R}} = \frac{1}{{\mathrm{R}}_1}+\frac{1}{{\mathrm{R}}_2}+\frac{1}{{\mathrm{R}}_3} \\ = \frac{1}{5}+\frac{1}{10}+\frac{1}{30} \\ = \frac{10}{30} \\ = \frac{1}{3} \end{gathered}$$

          $\mathrm{R} = 3 \mathrm{\Omega }.$