Question

Two electric bulbs A and B rated 200V~100W and 200V~60W are connected in series to a 200V line.Then the potential drop across
1)Each bulb is 200V
2)100W bulb is greater than that across 60W bulb
3)100W bulb is smaller than that across 60W bulb
4)Each bulb is 100V

Answer 1

The potential drop across 100W bulb is smaller than that across 60W bulb.

First bulb rating is 200 V - 100 W. So, the resistance of the first bulb is,

P1=V21R1 ⇒ R1=V21P1=2002100=40000100⇒ R1=400 ΩP1=V12R1 ⇒ R1=V12P1=2002100=40000100⇒ R1=400 Ω

Second bulb ratings is 200 V - 60 W. Therefore,

P2=V22R2 ⇒ R2=V22P2=200260=4000060⇒ R2=20003 ΩP2=V22R2 ⇒ R2=V22P2=200260=4000060⇒ R2=20003 Ω

Therefore, when two bulbs are connected in series, the effective resistance is,

Reff=R1+R2 ⇒ Reff=400+20003=1200+20003⇒ Reff=32003 ΩReff=R1+R2 ⇒ Reff=400+20003=1200+20003⇒ Reff=32003 Ω

So, current in the circuit is,

i=VReff=20032003/=6003200⇒ i=316 Ai=VReff=20032003=6003200⇒ i=316 A

The potential drop across each bulb is,

across 100 W bulb is,

V1=iR1=316×400=3×25⇒ V1=75 VV1=iR1=316×400=3×25⇒ V1=75 V

across 60 W bulb is,

V2=iR2=316×20003⇒ V2=125 VV2=iR2=316×20003⇒ V2=125 V

So, the potential drop across 100 W bulb is less than that of the potential drop across 60 W.