9.1 The simplified sketch of an electric motor is shown below - NSC Physical Sciences - Question 9 - 2022 - Paper 1

The motor shown in the figure is a DC motor, as indicated by the presence of a commutator, which is essential for reversing the direction of current in the coil and thus maintaining rotation.

The commutator in this motor functions to ensure continuous rotation of the coil. It reverses the direction of the current flowing through the coil as the motor turns, allowing the coil to experience a constant torque in the same direction.

To calculate the resistance of resistor Y, we use the formula for power:

$P = \frac{V^2}{R}$

Rearranging gives:

$R = \frac{V^2}{P}$

Substituting the values:

$R = \frac{220^2}{100} = \frac{48400}{100} = 484 \Omega$

Given that the power dissipated by resistor Y changes to 80 W, we can find the new current through resistor Y:

$P_{Y} = I_{Y}^2 R_{Y}$

Where: $I_{Y} = \sqrt{\frac{P_{Y}}{R_{Y}}} \Rightarrow I_{Y} = \sqrt{\frac{80}{484}} \approx 0.407 A$

Now, we calculate the voltage across resistor Y:

$V_{Y} = I_{Y} \times R_{Y} = 0.407 A \times 484 \Omega \approx 196.77 V$

Using the source voltage and the voltage across resistor Y, we can find the voltage across resistor Z:

$V_{Z} = 220 V - V_{Y} \approx 220 V - 196.77 V = 23.23 V$

Next, we calculate the current through resistor Z which is the same as the current through Y:

$I_{Z} = I_{Y} = 0.407 A$

The power rating of resistor Z is thus:

$P_{Z} = I_{Z}^2 R_{Z} = (0.407)^2 \times \frac{23.23}{R_{Z}}$

Assuming constant resistance and values from previous calculations, we can deduce:

$P_{Z} \approx 57.08 W$