Chapter 21, Problem 76GP

To determine

Conceptual states that an overloaded motor may burn may burn out due to high currents.

Answer to Problem 76GP

Solution:

The initial current through the blender is $40A$.

The back emf of the blender is $114V$.

The rate at which heat is dissipated in the blender is $12W$.

The rate at which heat is dissipated is $4800W$

Explanation of Solution

Given:

Formula used:

a) The initial current through the blender.:

$I=\frac{V}{R}$

Where, $I$ is the current through the blender, $V$ is the voltage, and $R$ is resistance of the blender

Substitute $120V$ for $V$ and $3.0\Omega forR.$

$\begin{array}{l}I=\frac{120V}{3.0\Omega }\\ =40A\end{array}$

b)The back emf of the blender:

Back emf from the normal operating current is,

$\begin{array}{l}{\xi }_{applied}-{\xi }_{back}=IR\\ {\xi }_{back}={\xi }_{applied}-IR\end{array}$

Where, ${\xi }_{back}$ is back emf, and ${\xi }_{applied}$ is applied emf.

Substitute $120V$ for ${\xi }_{applied}$, 2.0A for $I,$ and $3.0\Omega$ for $R.$

$\begin{array}{l}{\xi }_{back}=\left(120V\right)-\left(2.0A\right)\left(3.0\Omega \right)\\ =114V\end{array}$

c)The rate at which heat is dissipated in the blender.:

$\begin{array}{l}P=I^{2}R\\ \text{substitute}2.0A\text{for}I\text{and}3.0\Omega \text{for}R.\\ P={\left(2.0A\right)}^2\left(3.0\Omega \right)\\ =12\text{W}\end{array}$

d)The rate at which heat is dissipated:

$\begin{array}{l}P=I^{2}R\\ \text{substitute}40A\text{for}I\text{and}3.0\Omega \text{for}R.\\ P={\left(40A\right)}^2\left(3.0\Omega \right)\\ =4800\text{W}\end{array}$