Vijaya connects three bulbs P, Q and R is series with a battery in two different ways using identical conducting wires

Question

(a) Vijaya connects three bulbs P, Q and R is series with a battery in two different ways using identical conducting wires as shown below. She notices that in case I all three bulbs glow but in case II only the bulbs P and R continue to glow. What could be the reason for the bulb Q to not glow in case II? Explain

(b) Two resistances when connected in parallel give a combined resistance of 10/3 ohms. When the same two resistors are connected in series, the combined resistance becomes 15 ohms. Calculate the individual resistance of each resistor.

Answer 1

(a) The current will flow through the additional wire that connects the points L and M (avoiding the bulb) as it offers a path of least/lower resistance as compared with the bulb.

(b) 3/10 = 1/R₁ + 1/R₂ ………………. (1) 

R₁ + R₂ = 15, R₁ = 15 – R₂ 

Substituting in (1) 

3/10 = (15 – R₂ + R₂)/ (15 – R₂) R₂ 

15R₂ – R₂² = 150/3 = 50 

R₂² - 15R₂ + 50 = 0 

R₂ = 10 ohm, R₁ = 5 ohm 

or R₁ = 10 ohm, R₂ = 5 ohm