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