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

Explanation:

Part a

Step1: Identify values

We have \( E_{\text{max}} = 240.0 \) volts and \( \theta = 140^\circ \). The formula is \( e = E_{\text{max}} \sin\theta \).

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

First, find the sine of \( 140^\circ \). \( 140^\circ \) is in the second quadrant, and \( \sin(140^\circ)=\sin(180^\circ - 40^\circ)=\sin(40^\circ)\approx0.6428 \).

Step3: Compute \( e \)

Substitute into the formula: \( e = 240.0\times\sin(140^\circ)\approx240.0\times0.6428 = 154.272 \). Round to the nearest tenth: \( 154.3 \) volts.

Part b

Step1: Identify values

Here, \( E_{\text{max}} = 600.0 \) volts and \( \theta = 235^\circ \). Use \( e = E_{\text{max}} \sin\theta \).

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

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

Step3: Compute \( e \)

Substitute: \( e = 600.0\times\sin(235^\circ)\approx600.0\times(- 0.8192)=-491.52 \). The instantaneous voltage's magnitude is considered here (but since it's alternating, the sign indicates direction; for the value to the nearest tenth, it's \( - 491.5 \) volts (or 491.5 volts in magnitude if direction is not considered, but following the formula, it's \( - 491.5 \)).

Part c

Step1: Identify values

\( E_{\text{max}} = 120.0 \) volts and \( \theta = 310^\circ \). Use \( e = E_{\text{max}} \sin\theta \).

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

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

Step3: Compute \( e \)

Substitute: \( e = 120.0\times\sin(310^\circ)\approx120.0\times(- 0.7660)=-91.92 \). Round to the nearest tenth: \( - 91.9 \) volts (or 91.9 volts in magnitude, but formula gives \( - 91.9 \)).

Part a Answer: \( 154.3 \) volts

Part b Answer: \( - 491.5 \) volts (or \( 491.5 \) volts if magnitude)

Part c Answer: \( - 91.9 \) volts (or \( 91.9 \) volts if magnitude)

(Note: If the problem expects the magnitude for instantaneous voltage in terms of just the value (ignoring sign for magnitude), adjust accordingly, but the formula includes the sign based on the quadrant of the angle.)

Answer:

Step1: Identify values

\( E_{\text{max}} = 120.0 \) volts and \( \theta = 310^\circ \). Use \( e = E_{\text{max}} \sin\theta \).

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

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

Step3: Compute \( e \)

Substitute: \( e = 120.0\times\sin(310^\circ)\approx120.0\times(- 0.7660)=-91.92 \). Round to the nearest tenth: \( - 91.9 \) volts (or 91.9 volts in magnitude, but formula gives \( - 91.9 \)).

Part a Answer: \( 154.3 \) volts

Part b Answer: \( - 491.5 \) volts (or \( 491.5 \) volts if magnitude)

Part c Answer: \( - 91.9 \) volts (or \( 91.9 \) volts if magnitude)

(Note: If the problem expects the magnitude for instantaneous voltage in terms of just the value (ignoring sign for magnitude), adjust accordingly, but the formula includes the sign based on the quadrant of the angle.)