Question

A 1250 Wradiant heater is constructed to operate at 115 V. (a) What is the current in the heater when the unit is operating? (b) What is the resistance of the heating coil? (c) How much thermal energy is produced in 1.0 h?

Short answer

1. The current in the heater when the unit is operating is 10.9 A.
2. The resistance of the heating coil is $10.6\Omega$.
3. The thermal energy produced in is $4.50\times {10}^{6}J$.

Step-by-step solution

Identification of given data

1. The power dissipated is P = 1250W.
2. The potential difference is V = 115V.
3. The time of energy production is t = 1.0h or 3600s.

Significance of the flow of current and energy

The power or rate of energy transfer, in an electrical device across which a potential difference is equal to the product of current and potential difference. It can also be defined as energy transferred per unit time.

By using the equation for the rate of electrical energy transferred, Ohm’s law, and the thermal energy, we can find the current in the heater when the unit is operating, the resistance of the heating coil, and the thermal energy.

Formulae:

The rate of the electrical energy transferred, P = iV …(i)

Here, P is the power, i is current, V is the potential difference.

According to Ohm’s law, the potential difference is given by, V = iR …(ii)

Here, R is the resistance

The thermal energy E is generated by the heater, E = Pt …(iii)

Here, t is the time.

(a) Determining the current in the heater

Using the given data in equation (i), we can get the value of the current flowing into the heater as follows:

$\begin{array}{l}i=\frac{1250W}{115V}\\ =10.9A\end{array}$

Hence, the value of the current is 10.9 A.

(b) Determining the resistance in the heating coil

Using the given data in equation (ii), we can get the value of the resistance in the heating coils as follows:

$\begin{array}{l}R=\frac{115V}{10.9A}\\ =10.6\Omega \end{array}$

Hence, the value of the resistance is $10.6\Omega$.

(c) Determining the thermal energy

The thermal energy E is generated by the heater in time t = 3600s is given using the data substituted in equation (iii) as follows:

$$\begin{gathered} E=1250W\times 3600s \\ =4.50\times {10}^{6}J \end{gathered}$$

Hence, the amount of produced thermal energy is $4.50\times {10}^{6}J$.