The lightbulb in the circuit diagram of (Figure 1) has a resistance of 1.4 . Con

Question

The lightbulb in the circuit diagram of (Figure 1) has a resistance of 1.4 . Consider the potential difference between pairs of points in the figure. Suppose that E = 4.0 V .

Part A

What is the magnitude of V12?

Express your answer with the appropriate units.

Part B

What is the magnitude of V23?

Part C

What is the magnitude of V34?

Express your answer with the appropriate units.

Part D

What is the magnitude of V12 if the bulb is removed from the socket (i.e. the circuit is not closed)?

Part E

What is the magnitude of V23 if the bulb is removed from the socket (i.e. the circuit is not closed)?

1 of 1The circuit is shown in the figure. A battery of electromotive force E is in series with a 2.0-ohm resistor, which is in series with a bulb. The bulb is connected in series with the battery. Also four points are marked on the circuit. Point 1 is between the battery and the resistor. Point 2 is between the resistor and the bulb. Points 3 and 4 are between the bulb and the battery. Point 3 is closer to the bulb. Point 4 is closer to the battery.

The lightbulb in the circuit diagram of (Figure 1) has a resistance of 1.4 . Consider the potential difference between pairs of points in the figure. Suppose that E = 4.0 V .

Part A

What is the magnitude of V12?

Express your answer with the appropriate units.

Part B

What is the magnitude of V23?

Part C

What is the magnitude of V34?

Express your answer with the appropriate units.

Part D

What is the magnitude of V12 if the bulb is removed from the socket (i.e. the circuit is not closed)?

Part E

What is the magnitude of V23 if the bulb is removed from the socket (i.e. the circuit is not closed)?

1 of 1The circuit is shown in the figure. A battery of electromotive force E is in series with a 2.0-ohm resistor, which is in series with a bulb. The bulb is connected in series with the battery. Also four points are marked on the circuit. Point 1 is between the battery and the resistor. Point 2 is between the resistor and the bulb. Points 3 and 4 are between the bulb and the battery. Point 3 is closer to the bulb. Point 4 is closer to the battery.

Explanation / Answer

a) V12= IR = (4/3.4) x 2 = 2.35 V
b) V23 = (4/3.4) x 1.4 = 1.65 V  
c) V34 = 0 V  

when bulb is removed current = 4/2 = 2 A
d) V12 = 4 V   
e) V23 = 0 V

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