(a)  Show that the reciprocal of the equivalent resistance of a group of resistances joined in parallel is equal to the sum of the reciprocals of the individual resistance with the help of a circuit diagram.  

(b) In an electric circuit two resistors of 10Ω each are joined in parallel to a 5V battery. Find  

the current drawn from the battery. 

(a) 

Let us consider three resistors connected in parallel with a battery as R₁, 𝑅₂, 𝑅₃ are shown in the figure. 

The potential difference across each resistor is the same as the applied voltage when resistors are connected in parallel. But the value of current across each resistor is not the same. 

Let 𝐼₁+ 𝐼₂ + 𝐼₃ is the current that passes through each resistor 𝑅₁, 𝑅₂, 𝑅₃ respectively. 

So, 𝐼 = 𝐼₁+ 𝐼 ₂ + 𝐼 ₃           ........... equation (i)

Let us say the effective resistance is 𝑅_p, then using the Ohm's law

𝐼 = 𝑉/𝑅_p                             ............. equation (ii)

Because voltage is same for resistors in parallel combination, 

${\mathrm{I}}_1={\mathrm{V}}_1/{\mathrm{R}}_{\mathrm{p}}, {\mathrm{I}}_2={\mathrm{V}}_2/{\mathrm{R}}_{\mathrm{p}}, {\mathrm{I}}_3={\mathrm{V}}_3/{\mathrm{R}}_{\mathrm{p}}$— equation (iii) 

From equation (i), (ii) and (iii), we get,  

$$V/R_p=V_1/R_1+V_2/R_2+V_3/R_3$$

$$=V(1/R_1+1/R_2+1/R_3)$$

$$\Rightarrow 1/R_p=1/R_1+1/R_2+1/R_3$$

Hence, the reciprocal of effective resistance in parallel combination is equal to the sum of reciprocals of all the individual resistance.  

 (b) We have been given two resistors each of resistance 10 Ω and potential difference$$V=5 V$$So,

$$R_1=10𝛺$$

$$R_2= 10𝛺$$

$$V = 5V$$

We know the equivalent resistance in parallel combination is,  

$$1/R_p=1/R_1+1/R_2$$

$$=1/10+1/10$$

$$R_p=5𝛺$$

$$I=V/R_p$$

$$\Rightarrow I=5 V/5 𝛺$$

$$\Rightarrow I=1 A$$