Question

(a) How can three resistors of resistances 2 Ohm ,3 Ohm and 6 Ohm be connected to give a total resistance of (i) 4 Ohm, (ii) 1 Ohm.
(b)When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resister​

Answer 1

Answer:

(a)  Given :-

R₁ = 2 Ω      

R₂ = 3 Ω          

R₃ = 6 Ω      

Solution :-

(a)  When R2 and R3 are connected in parallel with R₁ in series we get                

⇒ 1/R = 1/R₂ + 1/R₃              

⇒ 1/R = 1/3 + 1/6                

⇒ 1/R = 1/2                

⇒ R = 2 Ω            

⇒ Resistance in series = R + R₁                

⇒ Resistance in series = 2 + 2 = 4 Ω

Hence, the total resistance of the circuit is 4 Ω.      

(b)  When R₁, R₂, R₃ are connected in parallel we get                

⇒ 1/R = 1/R₁ + 1/R₂ + 1/R₃                

⇒ 1/R = 1/2 + 1/3 + 1/6                

⇒ 1/R = 1 Ω.

Hence, the net equivalent resistance of the circuit is 1 Ω.

(b)  Given: V=12 V

           I=2.5 mA

Let the resistance be R

By Ohm's Law,  

V=IR

12=2.5×10³R

R=4.8×10³Ω