The lightbulb in the circuit diagram of (Figure 1) has a resistance of 0.60 . Co

Question

The lightbulb in the circuit diagram of (Figure 1) has a resistance of 0.60 . Consider the potential difference between pairs of points in the figure. Suppose that E = 3.0 V .

Part A

What is the magnitude of V12?

Express your answer with the appropriate units.

Part B

What is the magnitude of V23?

Express your answer with the appropriate units.

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)?

Express your answer with the appropriate units.

Part E

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

Express your answer with the appropriate units.

Part F

What is the magnitude of V34 if the bulb is removed from the socket (i.e. the circuit is not closed)?

Express your answer with the appropriate units.

Explanation / Answer

0.60 ohms and 2 ohms are in series.

hence, total resistance in the circuit = 2+0.60 = 2.6 ohms

then current in the circuit = emf/resistance = 3/2.6 = 1.153 A

part A:

voltage difference across point 1 and 2 is voltage difference across the resistance 2 ohms

then V12 = resistance x current = 2 x 1.153 = 2.307 volts

part B:

voltage across 2 and 3 is the voltage difference across the light bulb i.e. voltage difference across 0.6 ohms

hence, V23 = resistance x current = 0.6 x 1.153 = 0.691 volts

part C:

point 3 and point 4 are connected by a resistance less wire. hence they are at the same potential.

hence potential difference, V34 = 0 volts

part D:

if the circuit is not closed, then there is no current flow.

hence voltage drop across the resistance = resistance x current = 0 volts

hence V12 = 0 volts

part E:

as 2 and 3 are open circuited,and there is no voltage drop across 1 and 2,

point 2 is at same potential as point 1 and point 3 as same potential as point 4.
hence, V23 = V14 = emf of the battery = 3 volts

part F:
point 3 and point 4 are connected by a resistance less wire. hence they are at the same potential

hence potential difference, V34 = 0 volts

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