Answer to Question #256267 in Electrical Engineering for Juswah

Question

6. A 72-ohm coil of wire is connected in series with an adjustable resistor(rheostat) whole resistance can be varied from 0 to 99 ohms. If the line is 115 volts, calculate (a) the range through which the coil current can be varied, (b) the rheostat resistance when the power taken by the coil is 90 watts.

 

The answer should be:0.72 to 1.6 amp

 

7. Using the given data of Prob. 6, calculate the power taken by the coil when the voltage drops across rheostat and coil are equal.

The answer should be: 45.8 watts

Answer 1

(a) The minimum current:

$$I_\text{max}=V/r=115/72=1.60\text{ A},\\ I_\text{min}=V/(r+R)=115/(72+99)=0.673\text{ A}.$$

(b) The power is 90 W with the following value of resistance:

$$P=\frac{V^2}{r+R_1},\\\space\\ R_1=\frac {V^2}P-r=75\Omega.$$

The voltage drops are equal when both resistances are equal:

$$P=\frac{V}{r+R_2}r=\frac{115}{72+72}72=57.5\text{ W}.$$