V=2.5 V

P=?

Energy, E = ?

I = 750 mA = 0.75 A

R=?

t = 4 hours

Its power

P=VI = 2.5×0.75 = 1.87 W

Its resistance

R=2.5/0.75 = 3.33 W

The energy consumed if this bulb is lighted for 4 hours

E=Pt = 7.5 Wh

In series

R = R1+R2 = 2+2 = 4Ω

I=V/R = 12/4 = 3A

P1 = VI = 36 W

In parallel

1/R = 1/R1+1/R2 = 1/2+1/2 = 1Ω

I=V/R = 12A

P2=VI = 12W

Hence, P1/P2 = 36/12 = 3/1 = 3:1

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