Four identical electrical lamps are labelled 1.5 V 0.5 A which describes the condition necessary for them to operate at normal brightness. If a 12 V battery of negligible internal resistance is connected to lamps as shown, then

A
the value of R for normal brightness of each lamp is $(3 / 4) Ω$
B
the value of R for normal brightness of each lamp is $(21 / 4) Ω$
C
total power dissipated in circuit when all lamps are normally bright is 24 W
D
power dissipated in R is 2l W when all lamps ate normally bright

Solution:

For normal brightness of each bulb see following circuit. Current through each bulb = 0.5 A

So main current i = 2A
Also, voltage across the combination = 1.5 V
So voltage across the resistance = 10.5 V
Hence. for resistance, $V = iR \Rightarrow 10 .5 = 2 \times R = R = \frac{21}{4} Ω$

Solution:

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