Question

1. compute the instantaneous voltage (e) to the nearest tenth volt of an alternating electromotive force (emf) for each of these problems. use the formula

$e = e_{max} \sin \theta$
where $e = $ instantaneous voltage
$e_{max} = $ maximum voltage
$\theta = $ angle in degrees

number of degrees reached in cycle ($\theta$) \tmaximum voltage ($e_{max}$) \tinstantaneous voltage ($e$)

a. \t$140^\circ$ \t240.0 volts \tshow answer

b. \t$235^\circ$ \t600.0 volts \tshow answer

c. \t$310^\circ$ \t120.0 volts \tshow answer

d. \t$180^\circ$ \t800.0 volts \tshow answer

e. \t$340^\circ$ \t300.00 volts \tshow answer

Explanation:

Step1: For part d (θ = 180°, \( E_{\text{max}} = 800.0 \) volts)

First, find \( \sin(180^\circ) \). We know from trigonometry that \( \sin(180^\circ) = 0 \).

Step2: Apply the formula \( e = E_{\text{max}} \sin \theta \)

Substitute \( E_{\text{max}} = 800.0 \) and \( \sin(180^\circ) = 0 \) into the formula: \( e = 800.0 \times 0 = 0.0 \) volts.

Step3: For part e (θ = 340°, \( E_{\text{max}} = 300.00 \) volts)

First, find \( \sin(340^\circ) \). Using a calculator, \( \sin(340^\circ) \approx \sin(360^\circ - 20^\circ) = -\sin(20^\circ) \approx -0.3420 \).

Step4: Apply the formula \( e = E_{\text{max}} \sin \theta \)

Substitute \( E_{\text{max}} = 300.00 \) and \( \sin(340^\circ) \approx -0.3420 \) into the formula: \( e = 300.00 \times (-0.3420) = -102.6 \) volts (rounded to the nearest tenth).

Answer:

• Part d: \( 0.0 \) volts
• Part e: \( -102.6 \) volts