Question

part a
if the current to a motor drops by 18%, by what factor does the output torque change?
express your answer usint two significant figures.
\\(\frac{\tau_2}{\tau_1}=\\)

Explanation:

Step1: Define initial and new current

Let the initial current be $I_1$ and the new current $I_2$. Given that the current drops by 18%, so $I_2=(1 - 0.18)I_1=0.82I_1$.

Step2: Recall torque - current relationship

For a DC - motor, the torque $\tau$ is proportional to the current $I$, i.e., $\tau = kI$ (where $k$ is a constant). Let the initial torque be $\tau_1 = kI_1$ and the new torque be $\tau_2 = kI_2$.

Step3: Find the ratio of torques

$\frac{\tau_2}{\tau_1}=\frac{kI_2}{kI_1}$. Substitute $I_2 = 0.82I_1$ into the ratio: $\frac{\tau_2}{\tau_1}=\frac{0.82I_1}{I_1}=0.82$.

Answer:

$0.82$