40. A torch bulb is rated 2.5 V and 750 mA. Calculate (i) its power, (ii) its resistance and (iii) the energy consumed, if this bulb is lighted for four hours.
Answers
Answer:
1.875W = 1875 mW
0.33 Ω
7.5 Wh or 27000 J
Explanation:
Using, P = Vi ; P = i²R ; E = Pt
Here,
V = 2.5 V ; i = 750mA = 0.75 A
(i): P = Vi = (2.5)(0.75)
= 1.875W = 1875 mW
(ii): R = V/i = 2.5/0.75
= 0.33Ω
(iii): E = Pt = (1.875W)(4hr)
= 7.5 Wh
E can also be written in terms of Joule:
E = 7.5 W(3600 sec) [1hr = 3600sec]
= 27000 J
Given :-
A torch bulb is rated 2.5 V and 750 mA.
To Find :-
(i) its power, (ii) its resistance and (iii) the energy consumed, if this bulb is lighted for four hours.
Solution :-
(i) Power
1 A = 1000 mA
750 mA = 750/100 = 0.75 A
Power = Potential difference × Current
P = 0.75 × 2.5
P = 1.875 W
(ii) Resistance
Resistance = Voltage/Current
R = 2.5/0.75
R = 25/75 × 100/10
R = 1/3 × 10
R = 3.33 Ω
(iii) Energy
Energy = Power × Time
1 hr = 3600 sec
4 hr = 4 × 3600
4 hr = 14400 sec
Energy = 1.875 × 14400
Energy = 27000J