Question

A 85 W electric blanket runs at 24 V .

What is the resistance of the wire in the blanket?
How much current does the wire carry?

 

Answer 1

Concepts and reason

The concepts used to solve this problem are resistance and current through the wire. Initially, use the relation between power, voltage, and resistance to find the resistance of the wire in the blanket. Then, use the relation between current, voltage, and resistance to find the current through the wire in the blanket.

Fundamentals

Expression for the power in terms of voltage and resistance is, \(P=\frac{V^{2}}{R}\)

Here, \(P\) is the power, \(R\) is the resistance, and \(V\) is the voltage. The expression for the current is, \(I=\frac{V}{R}\)

Here, \(I\) is the current.

(1) Expression for the power in terms of voltage and resistance is,

\(P=\frac{V^{2}}{R}\)

Rearrange the above expression for \(R\). \(R=\frac{V^{2}}{P}\)

Substitute \(85 \mathrm{~W}\) for \(P\) and \(24 \mathrm{~V}\) for \(V\).

\(R=\frac{(24 \mathrm{~V})^{2}}{85 \mathrm{~W}}\)

\(=6.776 \Omega\)

\(=6.8 \Omega\)

Therefore, the resistance of the wire is \(6.8 \Omega\).

The resistance of the wire is directly proportional to the square of the voltage. The increases in the voltage will increase the resistance of the wire. The resistance of the wire is inversely proportional to power. The increases in the power will decrease the resistance of the wire.

(2) The expression for the current is, \(I=\frac{V}{R}\)

Substitute \(6.8 \Omega\) for \(R\) and \(24 \mathrm{~V}\) for \(V\).

\(I=\frac{24 \mathrm{~V}}{6.8 \Omega}\)

\(=3.529 \mathrm{~A}\)

\(=3.5 \mathrm{~A}\)

Therefore, the amount of current-carrying the wire is \(3.5 \mathrm{~A}\).

The amount of current in the wire is directly proportional to the voltage. The increases in the voltage will increase the current in the wire. The amount of current in the wire is inversely proportional to the resistance. If the resistance of the wire increases, the current through the wire will decrease.

Part 1 The resistance of the wire is 6.8Ω.

Part 2 The amount of current-carrying the wire is 3.5A.