two electric bulbs rated 200W, 220V and 100W, 220V are connected in parallel to an electric mains 220V.
Answers
Answer:
In case of Bulb 1
Potential difference,V1= 220v
Power,P1 = 100w
In case of Bulb 2
Potential Difffernce,D2 =220v
Power,P2 = 60w
Now, since they are connected in parallel combination hence they have same potential difffernce and the total Current in the circuit is given the sum of currents flowing in both the resistors.
Power = Potential difference×Current
=>P = V×I
=>I= P/V.
Now,In case of Bulb 1
I1 = 100/220
=> I1 = 5/11 watts -(i)
Now,in case 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.
Remember
1)Ohm's law of resistance
V=I×R
2) Power = Work/Time
or
Power = V×I = I^2×R
or there is an another step
Power = P = 100 W
Voltage = V = 220v
Resistance = R
P = V²/R
100 = 220 ×220/R
R = 220 × 220/100
= 484 Ω
Power = P = 60 W
Voltage = V = 220v
Resistance = R
P = V²/R
60 = 220 × 220/R
R = 220× 220/60
= 806.7 Ω
As the resistors are connected in parallel ,
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
The current drawn is 0.73 A