# A small motor draws a current of 1.76 A from a 110 V line. The output power of the motor is 0.20...

## Question:

A small motor draws a current of 1.76 A from a 110 V line. The output power of the motor is 0.20 hp.

(a) At a rate of 0.060/kWh, what is the cost, in cents, of operating the motor for 4.0 h? (b) What is the efficiency of the motor? ## Ohm's Law: Ohm's law relates the voltage across the conductor to the current flowing across it according to which the current {eq}(I) {/eq}flowing through the conductor is directly proportional to the voltage {eq}(V) {/eq} applied across the conductor. For the relation of Ohm's law: $$V=IR$$ where {eq}R{/eq} is the resistance of the conductor. ## Answer and Explanation: Given • Current flowing in the motor is {eq}I= 1.76\ \rm A {/eq} • Voltage across the motor is {eq}V=110\ \rm V {/eq} • The output power of the motor is {eq}P=0.20\ \rm hp {/eq} (a) The input power of the motor is given by: \begin{align} P_i&=VI \\[0.3cm] &=110(1.76) \\[0.3cm] &=193.6\ \rm W \end{align} Then, the energy used by the motor is: \begin{align} E&=P_i t\\[0.3cm] &=(193.6\ \rm W)(4\ \rm hr)\\[0.3cm] &=774.4\ \rm W \cdot h\\[0.3cm] &=0.774\ \rm kW \cdot h \end{align} Finally, the cost of operating the motor is: \begin{align} C&=0.774\ \rm kW \cdot h \left(0.060\ \frac{\}{\rm kW \cdot h} \right)\\[0.3cm] &\approx \ 0.046 \end{align}\\ Thus, the cost of operating the motor is {eq}\0.046 {/eq}.

(b)

To determine the efficiency of the motor, we must first solve for the power output in Watts.

\begin{align} P_o&=(0.20\ \rm hp)\left(\frac{746\ \rm W}{1\ \rm hp} \right)\\[0.3cm] &=149.2\ \rm W \end{align}

Now, the efficiency of the motor is equal to the ratio of power output to the power input:

\begin{align} \eta&=\frac{P_o}{P_i}\times 100\\[0.3cm] &=\frac{149.2}{193.6}\times 100\\[0.3cm] &\approx 77 \% \end{align} \\

Therefore, the efficiency of the motor is {eq}77 \%{/eq}.