Exam 27: Circuits

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Here is a loop equation: . What does this equation represent?

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In the figure, voltmeter V1 reads 600 V, voltmeter V2 reads 580 V, and ammeter A reads 100 A. The power wasted in the transmission line connecting the power house to the consumer is: In the figure, voltmeter V<sub>1</sub> reads 600 V, voltmeter V<sub>2</sub> reads 580 V, and ammeter A reads 100 A. The power wasted in the transmission line connecting the power house to the consumer is:

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Resistor 1 has twice the resistance of resistor 2. They are connected in parallel to a battery. The ratio of the thermal energy dissipation by 1 to that by 2 is:

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When switch S is open, the ammeter in the circuit shown reads 2.0 A. When S is closed, the ammeter reading: When switch S is open, the ammeter in the circuit shown reads 2.0 A. When S is closed, the ammeter reading:

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A battery of emf 24 V is connected to a 6.0- Ω\Omega resistor. As a result, current of 3.0 A exists in the resistor. The rate at which energy is being dissipated in the battery is:

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An ideal battery has an emf of 12 V. If it is connected to a circuit and creates a current of 4.0 A, what is the power?

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A galvanometer has an internal resistance of 12 Ω\Omega and requires 0.01 A for full scale deflection. To convert it to a voltmeter reading 3 V full scale, one must use a series resistance of:

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In the diagrams, all light bulbs are identical and all emf devices are identical. In which circuit (I, II, III, IV, V) will the bulbs be dimmest? In the diagrams, all light bulbs are identical and all emf devices are identical. In which circuit (I, II, III, IV, V) will the bulbs be dimmest?

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If a circuit has L closed loops, B branches, and J junctions the number of independent loop equations is:

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Resistances of 2.0 Ω\Omega , 4.0 Ω\Omega , and 6.0 Ω\Omega and a 24-V battery are all in series. The current in the 2.0 Ω\Omega resistor is:

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Each of the resistors in the diagram is 12 Ω\Omega . The resistance of the entire circuit is:  Each of the resistors in the diagram is 12   \Omega . The resistance of the entire circuit is:

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In the circuit shown, the capacitor is initially uncharged, and V = 10 V. At time t = 0, switch S is closed. If τ \tau denotes the time constant, the approximate current through the 3 Ω\Omega resistor when t = τ \tau /10 is:  In the circuit shown, the capacitor is initially uncharged, and V = 10 V. At time t = 0, switch S is closed. If    \tau  denotes the time constant, the approximate current through the 3   \Omega   resistor when t =    \tau  /10 is:

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A battery with an emf of 12 V and an internal resistance of 1 Ω\Omega is used to charge a battery with an emf of 10 V and an internal resistance of 1 Ω\Omega . The current in the circuit is:

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In the diagram R1 > R2 > R3. Rank the three resistors according to the current in them, least to greatest. In the diagram R<sub>1</sub> > R<sub>2</sub> > R<sub>3</sub>. Rank the three resistors according to the current in them, least to greatest.

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Resistances of 2.0 Ω\Omega , 4.0 Ω\Omega and 6.0 Ω\Omega and a 24-V emf device are all in series. The potential difference across the 2.0- Ω\Omega resistor is:

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Resistances of 2.0 Ω\Omega , 4.0 Ω\Omega , and 6.0 Ω\Omega and a 24-V emf device are all in series. The circuit is initially ungrounded. After grounding, the current in the circuit:

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In the circuit shown, both resistors have the same value R. Suppose switch S is initially closed and capacitor C is charged. When switch S is then opened, the circuit has a time constant τ \tau a. Conversely, suppose S is initially open and capacitor C is uncharged. When switch S is then closed, the circuit has a time constant τ \tau b. The ratio τ \tau a/ τ \tau b is:  In the circuit shown, both resistors have the same value R. Suppose switch S is initially closed and capacitor C is charged. When switch S is then opened, the circuit has a time constant    \tau  <sub>a</sub>. Conversely, suppose S is initially open and capacitor C is uncharged. When switch S is then closed, the circuit has a time constant    \tau  <sub>b</sub>. The ratio    \tau  <sub>a</sub>/   \tau  <sub>b</sub> is:

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In an antique automobile, a 6-V battery supplies a total of 48 W to two identical headlights in parallel. The resistance of each bulb is:

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A certain capacitor, in series with a resistor, is being charged. At the end of 10 ms its charge is half the final value. The time constant for the process is about:

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A total resistance of 3.0 ω \omega is to be produced by combining an unknown resistor R with a 12 ω \omega resistor. What is the value of R and how is it to be connected to the 12 Ω\Omega resistor?

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