two electric bulbs rated 100W;220V and 60W;220V are connected in parallel to an electric mains of 220V. Find the current drawn by the bulbs from the mains
Answers
Answer:
0.73 A
Explanation:
Power of first bulb = P = 100 W (Given)
Voltage of first bulb = V = 220v (Given)
Power of second bulb = P = 60 W (Given)
Voltage of second bulb = V = 220v (Given)
Resistance = R
P = V²/R ( first bulb)
100 = 220 ×220/R
R = 220 × 220/100
= 484 Ω
P = V²/R ( second bulb)
60 = 220 × 220/R
R = 220× 220/60
= 806.7 Ω
As the resistors are connected in parallel , thus, the total resistance = 1/R
1/R = 1/484 + 1/806.7
= 806.7 + 484/484×806.7
= 1290.7/390442.8
R = 390442.8/ 1290.7 = 302.5 ohms
Total resistance = 302.5 Ω
I = current
V = IR
220 = I × 302.5
I = 220/302.5
= 0.73 A
In the illustration of Bulb 1
Potential difference,V1= 220v
Power, P1 = 100w
In the illustration of Bulb 2
Potential Difffernce,D2 =220v
Power,P2 = 60w
Because they are attached in parallel succession hence they have the identical potential difference and the total Current in the line is granted the sum of currents moving in both the resistors.
Power = Potential difference×Current
P = V×I
I= P/V.
Now,In the situation of Bulb 1
I1 = 100/220
I1 = 5/11 watts ------------(i)
Now,in scenario if bulb 2
I2 = 60/220
I2= 3/11watts ---------------(ii)
From (i) and (ii) we get
Total Current in circuit,(I)n=>
(I)1 + (I)2
(3/11)+(5/11)
8/11 watts.
Hence, the current in the circuit is 8/11 watts or 0.73 watts.