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 ($\theta$)	maximum voltage ($e_{max}$)	instantaneous voltage ($e$)
b.	$235^\circ$	$600.0$ volts	show answer
c.	$310^\circ$	$120.0$ volts	show answer

$\tan = \frac{opp}{adj}$
$\csc = \frac{hyp}{opp}$

Explanation:

Part a

Step1: Identify values

$\theta = 140^\circ$, $E_{max} = 240.0$ volts, formula $e = E_{max} \sin\theta$.

Step2: Calculate $\sin(140^\circ)$

$\sin(140^\circ)=\sin(180^\circ - 40^\circ)=\sin(40^\circ)\approx0.6428$ (using reference angle).

Step3: Compute $e$

$e = 240.0\times0.6428\approx154.3$ volts.

Part b

Step1: Identify values

$\theta = 235^\circ$, $E_{max} = 600.0$ volts, formula $e = E_{max} \sin\theta$.

Step2: Calculate $\sin(235^\circ)$

$235^\circ$ is in third quadrant, $\sin(235^\circ)=\sin(180^\circ + 55^\circ)=-\sin(55^\circ)\approx - 0.8192$.

Step3: Compute $e$

$e = 600.0\times(-0.8192)\approx - 491.5$ volts.

Part c

Step1: Identify values

$\theta = 310^\circ$, $E_{max} = 120.0$ volts, formula $e = E_{max} \sin\theta$.

Step2: Calculate $\sin(310^\circ)$

$310^\circ$ is in fourth quadrant, $\sin(310^\circ)=\sin(360^\circ - 50^\circ)=-\sin(50^\circ)\approx - 0.7660$.

Step3: Compute $e$

$e = 120.0\times(-0.7660)\approx - 91.9$ volts.

Answer:

a. $154.3$ volts
b. $-491.5$ volts
c. $-91.9$ volts