Exam 6: Series-Parallel Circuits

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One Ohm's law formula PT = P1 + P2 + etc. can be used to solve for total power in:

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  -If R4 shorts in Figure 6 -2, VR5 . -If R4 shorts in Figure 6 -2, VR5 .

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In the parallel portion of series -parallel circuits, the total resistance is:

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   -If Vs = 12 V in Figure 6 -3, what is the value of VEB if R3 shorts? -If Vs = 12 V in Figure 6 -3, what is the value of VEB if R3 shorts?

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Two or more resistors connected in series form a circuit known as a voltage divider.

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The Wheatstone bridge circuit is widely used to measure:

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Thevinizing a circuit creates an equivalent series circuit.

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  -In Figure 6 -2, R3 and R4 are connected in . -In Figure 6 -2, R3 and R4 are connected in .

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A voltmeter, when connected across a component, can be viewed as being a resistor in series with that component.

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In the series portion of series -parallel circuits, the total resistance is:

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Maximum power is achieved when the load resistance is approximately two times the source resistance.

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The super position theorem provides a method for:

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The difference between a balanced and an unbalanced Wheatstone bridge is measured by:

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   -If R1 is changed to 500 Ω, the Thevenin resistance and voltage would be: -If R1 is changed to 500 Ω, the Thevenin resistance and voltage would be:

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   -If Vs = 50 V in Figure 6 -3, what is the value of VCA? -If Vs = 50 V in Figure 6 -3, what is the value of VCA?

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The voltage across any open in a series -parallel circuit will be the source voltage.

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A bridge circuit's resistances must all be of the same value to be in a balanced condition.

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  -If Vs = 15 V and every resistor equals 2.2 kΩ in Figure 6 -2, what is the value of IR4? -If Vs = 15 V and every resistor equals 2.2 kΩ in Figure 6 -2, what is the value of IR4?

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The parallel combination of a 330 Ω resistor and a 470 Ω resistor is connected in series with the parallel combination of four 1 kΩ resistors. If a 100 V source is connected across the circuit, then which resistor carries the most current?

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In solving series -parallel circuits using Ohm's law, first solve for:

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