Question

Q.3) Solve the following
2. Two tungsten bulbs of wattage 100 W and 60 W power work on 220 V potential
difference. if they are connected in parallel, how much current will flow in the main
conductor?
Ans...​

Answer 1

✬ Current = 0.72 A ✬

Explanation:

Given:

• Power of first bulb is 100W.
• Power of second bulb is 60W.
• Potential difference is 220 V.
• Both bulb are connected in parallel.

To Find:

• How much current will flow in main conductor ?

Formula to be used:

• P = VI

Solution: Potential difference remains same in the resistors when they are connected in the parallel.

Let see the first bulb

• P1 = 100 W.
• V = 220 V
• I = ?

$\implies{\rm }$ P1 = 220 × I

$\implies{\rm }$ 100 = 220 × I

$\implies{\rm }$ 100/220 = I

$\implies{\rm }$ 0.45 A = I

• So current flowing in first bulb is 0.45 A.

Now, for the second bulb

• P2 = 60 W
• V = 220 V
• I = ?

$\implies{\rm }$ P2 = 220 × I

$\implies{\rm }$ 60 = 220 × I

$\implies{\rm }$ 60/220 = I

$\implies{\rm }$ 0.27 A = I

• So current flowing in second bulb is 0.27 A.

∴ Total current across the conductor will be the sum of current in both the bulbs.

➨ Total current = 0.45 + 0.27 = 0.72 A

Hence, 0.72 ampere current will flow in main conductor.

Answer 2

Answer:

8/11=(0.72A)

used formula is

Power= Voltage *Current

when is connected in llel the voltages are same =220v

for 100w bulb the current(i 1 )= 100/220 =5/11 A

for 60w bulb the current (i 2)=60/220 = 3/11 A

the total current i= i1 + i2 = (5/11)+(3/11) =8/11 A

or

the total power be100+60 =160W

v=220v

p=vi

i=(p/v)

i=160/220

i=8/11

i=0.72 A