Question

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 \( \alpha = 2 \times 10^{-4} / ^\circ \text{C} \))

• A. 1694\( ^\circ \)C
• B. 1235\( ^\circ \)C
• C. 694\( ^\circ \)C
• D. 1667\( ^\circ \)C

Answer 1

The Correct Option is A

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

Answer 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 \( T \):

\[ \frac{4}{3} - 1 = 2 \times 10^{-4} \times (T - 27) \]

\[ \frac{1}{3} = 2 \times 10^{-4} \times (T - 27) \]

\[ T - 27 = \frac{1}{3 \times 2 \times 10^{-4}} = 1667 \]

\[ T = 1667 + 27 = 1694^\circ \text{C} \]