A torch bulb rated as 4.5 W, 1.5 V is connected . . .

Question : A torch bulb rated as 4.5 W, 1.5 V is connected as shown in figure given below. 

(i) What should be the e.m.f. of the cell required to make this bulb glow at full intensity?
(a) 4.5 V 
(b) 1.5 V
(c) 2.67 V 
(d) 13.5 V

(ii) What is the current passing through 1Ω resistor?
(a) 4.5 A 
(b) 1.5 A
(c) 2.67 A 
(d) 13.5 A

Doubt by Mohini

Solution : 

r = 2.67 Ω


i) Method I 
For Bulb
PB = 4.5 W
VB = 1.5 V
PB = VB2/RB
RB = VB2/PB
RB = 1.5×1.5 / 4.5
  

RB = 0.5 Ω

Resistance of parallel resistance (RP) = 1Ω
Equivalent External Resistance (R) 

= RB.RP / (RB + RP)

= (0.5×1)/(0.5+1)
= 0.5/1.5
R = 1/3 Ω

We know, 
r = [ε/v - 1] R

r/R = ε/v - 1 
r/R + 1 = ε/v  
(r/R + 1)V = ε 
ε = (r/R + 1)V
ε = [2.67/(1/3) + 1]×1.5
ε = [(2.67×3) + 1]×1.5
ε = [ 8.01+1]×1.5
ε = 9.01×1.5
ε = 13.515
ε = 13.5 V
Hence, d) would be the correct option.


Method II 

For Bulb
PB = 4.5 W
VB = 1.5 V
Current through the bulb
PB=VB×IB
IB = PB/VB
IB = 4.5/1.5
IB = 3 A

Current through the 1Ω resistance which is connected in parallel
IP = VP/RP
IP= 1.5/1
IP = 1.5 A

Total current drawn from the battery 
I = IB+IP
I = 3+1.5
I = 4.5 A

We know, 
ε = v+Ir
ε = 1.5+4.5(2.67)
ε = 1.5+12.015
ε = 13.515
ε = 13.5 V
Hence, d) would be the correct option.

ii) Current passing through the 1Ω resistor
The voltage across the resistance is 1.5 V because it is connected in parallel. 
VP=IPRP
IP=VP/RP
IP=1.5/1
IP = 1.5 A
Hence, b) would be the correct option.