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

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(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.

by (227k points)

(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 - 15R+ 50 = 0

R2 = 10 ohm, R1 = 5 ohm

or R1 = 10 ohm, R2 = 5 ohm