1. 16.3 A
To calculate the average rms current, we have to calculate the average rms current first. The peak voltage of the power line is Vp = 120 V, so the rms voltage is
[tex]V_{rms}=\frac{V_p}{\sqrt{2}}=\frac{120 V}{\sqrt{2}}=84.9 V[/tex]
The family uses 1000 kWh of energy per month. Since
1 kW = 1000 W
1 h = 3600 s
This energy corresponds to
[tex]E=1000 kWh \cdot (1000 W/kW)\cdot (3600 s/h)=3.6\cdot 10^9 Ws=3.6\cdot 10^9 J[/tex]
The time interval considered is 1 month, which corresponds to
[tex]t=1 m = (30 d)(24 h)(60 min)(60 s)=2.6\cdot 10^6 s[/tex]
And the power is the ratio between the energy used and the time taken:
[tex]P=\frac{E}{t}=\frac{3.6\cdot 10^9 J}{2.6\cdot 10^6 s}=1385 W[/tex]
And now we can find the average rms current by using the relationship between power, voltage and current:
[tex]I_{rms}=\frac{P}{V_{rms}}=\frac{1385 W}{84.9 V}=16.3 A[/tex]
2. [tex]5.2 \Omega[/tex]
The average resistance of the household can be calculated by using Ohm's law. Using the value found for the rms voltage and the rms current, we find
[tex]R=\frac{V_{rms}}{I_{rms}}=\frac{84.9 V}{16.3 A}=5.2 \Omega[/tex]
Given the data in the question;
Electricity used; [tex]E = 1000kWh = 36*10^8J[/tex]
Time; [tex]t = 1 month = 2.592 * 10^6s[/tex]
Voltage; [tex]V = 120V[/tex]
Calculate the average rms current to the house.
First we determine the power
Power is the amount of energy transferred per unit time:
[tex]Power = \frac{E}{t}[/tex]
So we substitute in our values
[tex]Power = \frac{36*10^8J}{2.592*10^6s}\\\\Power = 1.388 * 10^3 J/s \\\\Power = 1.388 * 10^3 W[/tex]
Next we Calculate the Current
From Ohms Law:
[tex]P = I * V[/tex]
We substitute in our values
[tex]1.388*10^3W = I \ *\ 120V \\\\I = \frac{1.388*10^3W}{120V} \\\\I = 11.57 W/V\\\\I = 12A[/tex]
Therefore, the average rms current in the power line to the house is 12A
The resistance of a household
From Ohm's Law:
[tex]R = \frac{V}{I}[/tex]
We substitute in our values
[tex]R = \frac{120V}{12A}\\\\R = 10ohms[/tex]
Therefore, The average resistance of a household is 10Ω.
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