The lightbulb in the circuit diagram of (Figure 1) has a resistance of 0.90 . Co

Question

The lightbulb in the circuit diagram of (Figure 1) has a resistance of 0.90 . 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?

Part B

What is the magnitude of V23?

Part C

What is the magnitude of V34?

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)?

Part F

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

Explanation / Answer

given

E = 4 V , R = 2 Ohm , r = 0.9

i = E / R + r

i = 4 / ( 2 + 0.9 ) = 1.379 A

Part A )

V12 = i R

= 1.379 X 2

V12 = 2.758 volts

Part B )

V23 = i r

= 1.379 X 0.9

V23 = 1.2411 volts

Part C )

V34 = i R

= 1.379 X 0

= 0

Part D )

the magnitude of V12 if the bulb os

V12 = i X R

= 0

Part E )

V23 = E

= 4 V

Part F )

the magnitude of V34 if the bulb is

V34 = i R

= 0 X 0

= 0

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