Question

A typical American family uses $1000kWh$of electricity a month.

a. What is the average current in the $120V$power line to the house?

b. On average, what is the resistance of a household?

Short answer

(a) $I=11.6A$.

(b)$R=10.4\Omega$

Step-by-step solution

Part (a) step 1: Given information

We need to find the average current in the $120V$power line to the house.

Part (a) step 2: Simplify

The energy is dissipated when the current flows through a resistor. The rate of the dissipated energy is the power. It is given as

$P_R=I∆V_R$

Solving equation (1) to get current,

$I=\frac{P_R}{∆V_R}$

The $\left(kWh\right)$is a unit of energy, so to determine the power we calculate the rate of consumption per month. The power is

$P_R=1000kWh/month$

We must convert the unit into $\left(W\right)$as next

$P_R=1000\frac{kWh}{month}\frac{1000W}{1kW}\frac{1month}{30days}\frac{1day}{24h}=1389W$

Now, putting the values for $P_R$and $∆V_R$into equation (2) to get $I$by

$$\begin{gathered} I=\frac{P_R}{∆V_R} \\ =\frac{1389W}{120V} \\ =11.6A \end{gathered}$$

Part (b) step 1: Given information

We need to find the resistance of a household on average.

Part (b) step 2: Simplify

Using Ohm's law $∆V_R=IR$, an alternative formula for the dissipated power is given as

$P_R=\frac{{\left(∆V_R\right)}^2}{R}$

Rearrange equation (3) for $R$to be in the form

$R=\frac{{\left(∆V_R\right)}^2}{P_R}$

Putting the values $P_R$and $∆V_R$into equation (4) to get $R$

$$\begin{gathered} R=\frac{{\left(∆V_R\right)}^2}{P_R} \\ =\frac{{\left(120V\right)}^2}{1389W} \\ =10.4\Omega \end{gathered}$$