An electric toaster has resistance of \( 60 \, \Omega \) at room temperature \( (27^\circ \text{C}) \). The toaster is connected to a 220 V supply. If the current flowing through it reaches 2.75 A, the temperature attained by the toaster is around: (if \( \alpha = 2 \times 10^{-4} / ^\circ \text{C} \))
- 1694\( ^\circ \)C
- 1235\( ^\circ \)C
- 694\( ^\circ \)C
- 1667\( ^\circ \)C
The Correct Option is A
Approach Solution - 1
To solve this problem, we need to find the temperature attained by the toaster when it is connected to a 220 V supply, and the current flowing through it is 2.75 A. The given parameters are:
- Initial resistance at room temperature (\(27^\circ \text{C}\)): \( R_0 = 60 \, \Omega \)
- Voltage supply: \( V = 220 \, \text{V} \)
- Current: \( I = 2.75 \, \text{A} \)
- Temperature coefficient of resistance: \( \alpha = 2 \times 10^{-4} / ^\circ \text{C} \)
The resistance at the temperature the toaster finally reaches can be calculated using Ohm's Law:
\(R = \frac{V}{I}\)
Substituting the given values:
\(R = \frac{220}{2.75} = 80 \, \Omega\)
Let the final temperature be \( T \). The relation between resistance and temperature is given by:
\(R = R_0(1 + \alpha(T - T_0))\)
Substituting the known values:
\(80 = 60 \times (1 + 2 \times 10^{-4} \times (T - 27))\)
Solve for \( T \):
\(1.3333 = 1 + 2 \times 10^{-4} \times (T - 27)\)
\(0.3333 = 2 \times 10^{-4} \times (T - 27)\)
\(T - 27 = \frac{0.3333}{2 \times 10^{-4}}\)
\(T - 27 = 1666.5\)
\(T = 1666.5 + 27 = 1693.5\)
Hence, the temperature attained by the toaster is approximately \( 1694^\circ \text{C} \).
The correct answer is: 1694\( ^\circ \)C
Approach Solution -2
Calculate Resistance at Operating Temperature:
Given \( V = 220 \, \text{V} \) and \( I = 2.75 \, \text{A} \), use Ohm’s law to find the resistance at the elevated temperature:
\[ R = \frac{V}{I} = \frac{220}{2.75} = 80 \, \Omega \]
Use Temperature Coefficient of Resistance Formula:
The relation between the resistance at room temperature \( R_0 \) and the resistance at temperature \( T \) is given by:
\[ R = R_0 (1 + \alpha \Delta T) \]
Substitute \( R = 80 \, \Omega \), \( R_0 = 60 \, \Omega \), and \(\alpha = 2 \times 10^{-4} \ ^\circ \text{C}^{-1}\), where \( \Delta T = T - 27 \):
\[ 80 = 60 \left(1 + 2 \times 10^{-4} \times (T - 27)\right) \]
Solve for \( T \):
- Divide both sides by 60:
\[ \frac{80}{60} = 1 + 2 \times 10^{-4} \times (T - 27) \]
- Simplify and isolate \( T \):
\[ \frac{4}{3} - 1 = 2 \times 10^{-4} \times (T - 27) \]
\[ \frac{1}{3} = 2 \times 10^{-4} \times (T - 27) \]
\[ T - 27 = \frac{1}{3 \times 2 \times 10^{-4}} = 1667 \]
\[ T = 1667 + 27 = 1694^\circ \text{C} \]
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Concepts Used:
Resistance
Resistance is the measure of opposition applied by any object to the flow of electric current. A resistor is an electronic constituent that is used in the circuit with the purpose of offering that specific amount of resistance.
R=V/I
In this case,
v = Voltage across its ends
I = Current flowing through it
All materials resist current flow to some degree. They fall into one of two broad categories:
- Conductors: Materials that offer very little resistance where electrons can move easily. Examples: silver, copper, gold and aluminum.
- Insulators: Materials that present high resistance and restrict the flow of electrons. Examples: Rubber, paper, glass, wood and plastic.
Resistance measurements are normally taken to indicate the condition of a component or a circuit.
- The higher the resistance, the lower the current flow. If abnormally high, one possible cause (among many) could be damaged conductors due to burning or corrosion. All conductors give off some degree of heat, so overheating is an issue often associated with resistance.
- The lower the resistance, the higher the current flow. Possible causes: insulators damaged by moisture or overheating.