A torch bulb rated as 4.5 W, 1.5 V is connected as shown in fig. The e.m.f. of the cell, needed to make the bulb glow at full intensity is(a) 4.5 V(b) 1.5 V(c) 2.67 V(d) 13.5 V
Answers
As, Power of bulb is defined as = V * I
where, V = voltage rating of bulb = 1.5 V
I = current rating of bulb
Here, Power rating of bulb = 4.5 W
=> 4.5 = 1.5 * I
=> current rating of bulb , I = 4.5/1.5
= 3 A
In the circuit diagram , we observe that current in the branch where bulb is located is 2E/9 .
Applying Kirchoff's Law in the branch we find that -
=> 2E/9 = 3 A
=> 2E = 27
=> E = 27/2
= 13.5 V
Thus, The e.m.f. of the cell, needed to make the bulb glow at full intensity is 13.5 V .
So, correct option is (d) 13.5 V .
A torch bulb rated as 4.5 W, 1.5 V is connected as shown in fig. The e.m.f. of the cell, needed to make the bulb glow at full intensity is(a) 4.5 V(b) 1.5 V(c) 2.67 V(d) 13.5 V
There is something wrong with Data
as per figure current shown in bulb is double then current shown in resistance 0.33
so Resistance of bulb should be = 0.33/2 = 0.165 Ω
But resistance of bulb = V²/P = 1.5²/4.5 = 0.5 Ω
Bulb Rated = 4.5 W , 1.5V
P = V²/R
Resistance of Bulb = V²/P = 1.5²/4.5 = 0.5 Ω
now 0.5 & 0.33 resistance are in parallel
so net resistance 1/R = 1/0.5 + 1/0.33
1/R = 2 + 3
1/R = 5
R = 1/5 = 0.2 Ω
Resistance of Cell = 2.67 Ω
Total Resistance = 2.67 + 0.2 = 2.87 Ω
Current drawn from source = E / 2.87
Let say current drawn from bulb = I1 & from 0.33Ω = I2
I1 * 0.5 = I2 * 0.33
I2 = I1* 1.5
I1 + I.5I1 = I
2.5I1 = I
I1 = I/2.5
I1 = 0.4I
I1 = 0.4 (E / 2.87)
I1 = E/7.175
I1²Rbulb = 4.5
=> (E/7.175)² * 0.5 = 4.5
=> (E/7.175)² = 9
=> (E/7.175) = 3
=> E = 3 * 7.175
=> E = 21.525 V