“WAKE UP AND SMELL THE (TRANSFORMER)!”
A friend returns to the United States from Europe with a 960-W coffeemaker, designed to operate from a 240-V line. (a) What can she do to operate it at the USA-standard 120 V? (b) What current will the coffeemaker draw from the 120-V line? (c) What is the resistance of the coffeemaker? (The voltages are rms values.)
IDENTIFY and SET UP:
Our friend needs a step-up transformer to convert 120-V ac to the 240-V ac that the coffeemaker requires. We use Eq. (31.35) to determine the transformer turns ratio N_{2} / N_{1}, P_{\mathrm{av}}=V_{\mathrm{rms}} I_{\mathrm{rms}} for a resistor to find the current draw, and Eq. (31.37) to calculate the resistance.
\frac{V_{1}}{I_{1}}=\frac{R}{\left(N_{2} / N_{1}\right)^{2}} (31.37)
EXECUTE:
(a) To get V_2 = 240 V from V_1 = 120 V, the required turns ratio is N_{2} / N_{1}=V_{2} / V_{1}=(240 \mathrm{~V}) /(120 \mathrm{~V})=2. That is, the secondary coil (connected to the coffeemaker) should have twice as many turns as the primary coil (connected to the 120-V line).
(b) We find the rms current I_1 in the 120-V primary by using P_{\mathrm{av}}=V_{1} I_{1}, where P_{av} is the average power drawn by the coffeemaker and hence the power supplied by the 120-V line. (We’re assuming that no energy is lost in the transformer.) Hence I_1[= P_{\mathrm{av}} / V_{1}=(960 \mathrm{~W}) /(120 \mathrm{~V})=8.0 \mathrm{~A}. The secondary current is then I_{2}=P_{\mathrm{av}} / V_{2}=(960 \mathrm{~W}) /(240 \mathrm{~V})=4.0 \mathrm{~A}.
(c) We have V_{1}=120 \mathrm{~V}, I_{1}=8.0 \mathrm{~A}, \text { and } N_{2} / N_{1}=2, so
\frac{V_{1}}{I_{1}}=\frac{120 \mathrm{~V}}{8.0 \mathrm{~A}}=15 \OmegaFrom Eq. (31.37),
R=2^{2}(15 \Omega)=60 \Omega
EVALUATE: As a check, V_{2} / R=(240 \mathrm{~V}) /(60 \Omega)=4.0 \mathrm{~A}=I_{2}, the same value obtained previously. You can also check this result for R by using the expression P_{\mathrm{av}}=V_{2}^{2} / R for the power drawn by the coffeemaker.