Question

Question: A $\mathbf{540}-\mathbf{W}$electric heater is designed to operate from$\mathbf{120}-\mathbf{V}$lines. (a) What is its operating resistance?(b) What current does it draw?(c) If the line voltage drops to 110 V what power does the heater take? (Assume that the resistance is constant. Actually, it will change because of the change in temperature.)(d) The heater coils are metallic, so that the resistance of the heater decreases with decreasing temperature. If the change of resistance with temperature is taken into account, will the electrical power consumed by the heater be larger or smaller than what you calculated in part (c)? Explain.

Short answer

Answer

1. The operating resistance of the electric heater is $26.7\Omega$.
2. The current drawn by the electric heater is 4.50A.
3. If the line voltage drops to 110 V the heater take the power of 453W.

4. If the change of resistance with temperature is taken into account the electrical power consumed by the heater be larger than 453W .

Step-by-step solution

Define the ohm’s law, resistance and power 

According to Ohm’s law, the current flowing through the conductor is directly proportional to the voltage across the two points as:

V = IR

Where, I is current in ampere , R is resistance in ohms $\left(\Omega \right)$and Vis the potential difference volt (V).

If the real source of emf $\epsilon$has internal energy then its terminal potential difference$V_{ab}$ depends upon the current.

So, the formula for the terminal potential difference is:

$V_{ab}=\epsilon -Ir$

The ratio of to for a particular conductor is called its resistance R

$R=\frac{V}{I}or\frac{\rho L}{A}$

Where, $\rho$is resistivity $\Omega ·\text{m}$, L is the length in m and A is the area in ${\text{m}}^2$.

The power (P) is the product of potential difference (V) and the current (I) .

$$\begin{gathered} P=VI \\ \text{or} \\ P=I^{2}R \\ \text{or} \\ P=\frac{V^2}{R} \end{gathered}$$

Determine the operating resistance of the electric heater

a)

Given that,

$$\begin{gathered} P=540\text{W} \\ V=120\text{V} \end{gathered}$$

The resistance of electric heater:

$$\begin{gathered} R=\frac{V^2}{P} \\ =\frac{{120}^2}{540} \\ =26.7\Omega \end{gathered}$$

Hence, the operating resistance of the electric heater is$26.7\Omega$ .

 Step 3: Determine the current

b)

The current of the heater is:

$$\begin{gathered} I=\sqrt{\frac{P}{R}} \\ =\sqrt{\frac{540}{26.7}} \\ =4.50\text{A} \end{gathered}$$

Hence, the current drawn by the electric heater is 4.50A .

Determine power

c)

The power of the heater:

$$\begin{gathered} P=\frac{V^2}{R} \\ =\frac{{110}^2}{26.7} \\ =453\text{W} \end{gathered}$$

The power will be larger than 453W .

Determine if the change of resistance with temperature is taken into account, will the electrical power consumed by the heater be larger or smaller

d)

As according to expression $R=\frac{V^2}{P}$, the power of the heater is inversely proportional to the resistance. So, if the voltage of the heater drops, the temperature will decrease as there is no collide between the molecules in the electric heater, so the resistance of the heater will decrease. Therefore, the power will increase with a decrease in resistance and voltage drops.

Hence, if the line voltage drops to 110V the heater takes the power of 453W and if the change of resistance with temperature is taken into account the electrical power consumed by the heater be larger than 453W .