Respuesta :
Answer:
Therefore,
15 Ampere current is flowing in the line.
Explanation:
Given:
Voltage, V = 120 V
Power of Appliances,
P1 = 100-W lightbulb,
P2 = 180-W television set,
P3 = 230-W desktop computer,
P4 = 1050-W toaster, and
P5 = 240-W refrigerator.
To Find:
Current flowing in the line, I = ?
Solution:
We have Power formula,
[tex]Power=Voltage\times Current\\P=V\times I[/tex]
For Household Voltage Remain Same and Current will differ and there will be a Parallel Connection for the Appliances,
Therefore,
1 .For lightbulb,
[tex]P_{1}=V\times I_{1}[/tex]
Substituting the values we get
[tex]100 = 120\times I_{1}\\I_{1}=0.8333\ Ampere[/tex]
2. Similarly for television,
[tex]P_{2}=V\times I_{2}[/tex]
Substituting the values we get
[tex]180= 120\times I_{2}\\I_{2}=1.5\ Ampere[/tex]
3. Similarly for desktop computer,
[tex]P_{3}=V\times I_{3}[/tex]
Substituting the values we get
[tex]230= 120\times I_{3}\\I_{3}=1.9166\ Ampere[/tex]
4. Similarly for toaster,
[tex]P_{4}=V\times I_{4}[/tex]
Substituting the values we get
[tex]1050= 120\times I_{4}\\I_{4}=8.75\ Ampere[/tex]
5. Similarly for refrigerator,
[tex]P_{5}=V\times I_{5}[/tex]
Substituting the values we get
[tex]240= 120\times I_{5}\\I_{5}=2\ Ampere[/tex]
In Parallel connection Total current 'I' is given as,
[tex]I=I_{1}+I_{2}+I_{3}+I_{4}+I_{5}[/tex]
Substituting the values we get
[tex]I=0.8333+1.5+1.9166+8.75+2=14.9999\approx 15\ Ampere[/tex]
Therefore,
15 Ampere current is flowing in the line.
The total current flowing in the line is 15A.
Electrical Power and Current
It is given that the supply voltage is V = 120 V
Now the power consumed by different components is:
P₁ = 100-W lightbulb,
P₂ = 180-W television set,
P₃ = 230-W desktop computer,
P₄ = 1050-W toaster, and
P₅ = 240-W refrigerator.
Now the relation between power, voltage, and current is given by;
P = V×I
I = P/V
Since the supply voltage is the same for all, the individual currents for different appliances can be calculated as follows:
[tex]I_1=\frac{100}{120}A=0.83\;A\\\\ I_2=\frac{180}{120}A=1.5\;A\\\\I_3=\frac{230}{120}A=1.92\;A\\\\I_4=\frac{1050}{120}A=8.75\;A\\\\I_5=\frac{240}{120}A=2\;A\\\\[/tex]
Therefore total current:
I = I₁ + I₂ + I₃ + I₄ + I₅
I = 15A
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