| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.68 |
| Score | 0% | 74% |
In a series circuit, which of the following is the same across all branches of the circuit?
resistance |
|
current |
|
conductance |
|
voltage |
A series circuit has only one path for current to flow. In a series circuit, current (I) is the same throughout the circuit and is equal to the total voltage (V) applied to the circuit divided by the total resistance (R) of the loads in the circuit. The sum of the voltage drops across each resistor in the circuit will equal the total voltage applied to the circuit.
| 40 Ω | |
| 80 Ω | |
| 240 Ω | |
| 88 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 = \( \frac{V}{I} \) = \( \frac{160}{2} \) = 80 Ω
Which of the following is a difference between a circuit breaker and a fuse?
a circuit breaker can be reused |
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a fuse is cheaper than a circuit breaker |
|
all of these |
|
a fuse responds more quickly than a circuit breaker |
Like fuses, circuit breakers stop current flow once it reaches a certain amount. They have the advantage of being reusable (fuses must be replaced when "blown") but respond more slowly to current surges and are more expensive than fuses.
Which of the following will increase the magnetic field produced by the electric current in a wire?
wrap the wire around a ceramic core |
|
construct the wire from conductive material |
|
wind the wire into a coil |
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construct the wire from insulative material |
A moving electric current produces a magnetic field proportional to the amount of current flow. This magnetic field can be made stronger by winding the wire into a coil and further enhanced if done around an iron containing (ferrous) core.
| 7.5 A | |
| 6.75 A | |
| 15 A | |
| 9.5 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 = \( \frac{V}{R} \) = \( \frac{75}{10} \) = 7.5 A