| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.72 |
| Score | 0% | 74% |
| 364 W | |
| 360 W | |
| 363 W | |
| 361.5 W |
| parallel | |
| series-parallel | |
| orthogonal | |
| series |
Connecting the 10 batteries in series multiplies their voltage while keeping their current the same yielding a 150V 20A configuration. Connecting the 10 batteries in parallel multiplies their current while keeping their voltage the same yieleding a 15V 200A configuration. Using a series-parallel connection, 5 batteries can be connected in series and 5 can be connected in parallel resulting in a 75V 100A configuration.
Which of the following is the formula for calculating electrical power?
P = IV |
|
\(P = {V \over I}\) |
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\(P = {I \over V}\) |
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P = I2V |
Electrical power is measured in watts (W) and is calculated by multiplying the voltage (V) applied to a circuit by the resulting current (I) that flows in the circuit: P = IV. In addition to measuring production capacity, power also measures the rate of energy consumption and many loads are rated for their consumption capacity. For example, a 60W lightbulb utilizes 60W of energy to produce the equivalent of 60W of heat and light energy.
A capacitor is often used for which of the following purposes?
as a primary power source |
|
to convert alternating current into direct current |
|
to allow a small amount of current to control a large amount of current |
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to maintain power in a circuit when the primary power source is disconnected |
Capacitors store electricity and are used in circuits as temporary batteries. Capacitors are charged by DC current (AC current passes through a capacitor) and that stored charge can later be dissipated into the circuit as needed. Capacitors are often used to maintain power within a system when it is disconnected from its primary power source or to smooth out or filter voltage within a circuit.
| 54 Ω | |
| 51 Ω | |
| 60 Ω | |
| 62 Ω |
Ohm's law specifies the relationship between voltage (V), current (I), and resistance (R) in an electrical circuit: V = IR.
Solved for resistance, R = \( \frac{V}{I} \) = \( \frac{90}{1.5} \) = 60 Ω