| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.51 |
| Score | 0% | 70% |
This circuit component symbol represents a(n):
capacitor |
|
resistor |
|
diode |
|
fuse |
Fuses are thin wires that melt when the current in a circuit exceeds a preset amount. They help prevent short circuits from damaging circuit components when an unusually large current is applied to the circuit, either through component failure or spikes in applied voltage.
Which of the following allows current to pass easily in one direction and blocks current in the other direction?
diode |
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inductor |
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capacitor |
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resistor |
A diode allows current to pass easily in one direction and blocks current in the other direction. Diodes are commonly used for rectification which is the conversion of alternating current (AC) into direct current (DC). Because a diode only allows current flow in one direction, it will pass either the upper or lower half of AC waves (half-wave rectification) creating pulsating DC. Multiple diodes can be connected together to utilize both halves of the AC signal in full-wave rectification.
The ampere is a unit of measurement for:
inductance |
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energy |
|
current |
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power |
Current is the rate of flow of electrons per unit time and is measured in amperes (A). A coulomb (C) is the quantity of electricity conveyed in one second by a current of one ampere.
| orthogonal | |
| parallel | |
| series-parallel | |
| perpendicular |
Connecting the 4 batteries in series multiplies their voltage while keeping their current the same yielding a 24V 5A configuration. Connecting the 4 batteries in parallel multiplies their current while keeping their voltage the same yieleding a 6V 20A configuration. Using a series-parallel connection, 2 batteries can be connected in series and 2 can be connected in parallel resulting in a 12V 10A configuration.
| 8.25 A | |
| 22.5 A | |
| 9.5 A | |
| 7.5 A |
Ohm's law specifies the relationship between voltage (V), current (I), and resistance (R) in an electrical circuit: V = IR.
Solved for current, I = \( \frac{V}{R} \) = \( \frac{150}{20} \) = 7.5 A