The lightbulb in the circuit diagram of (Figure 1) has a resistance of 1.2 . Con

Question

The lightbulb in the circuit diagram of (Figure 1) has a resistance of 1.2 . Consider the potential difference between pairs of points in the figure. Suppose thatE = 3.0 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

  E = 3.0 V
Rb = 1.2 ohm
R = 2.0 ohm
I = E/(Rb+R) = 0.9375 A

A.) What is the magnitude of V12?
V12 = 2*0.9375 =1.875 V

B.) What is the magnitude of V23?
V23 = 1.2*0.9375 = 1.125 V

C.) What is the magnitude of V34?
V23 = 0*0.9375 = 0 V

D.) What is the magnitude of V12 if the bulb is removed from the socket (i.e. the circuit is not closed)?
V12 = 2.0*0 = 0 V

E.) What is the magnitude of V23 if the bulb is removed from the socket (i.e. the circuit is not closed)?
V23 = E-V12 = 3.0-0 = 3.0 V

F.) What is the magnitude of V34 if the bulb is removed from the socket (i.e. the circuit is not closed)?
V34 = 0*0 = 0 V

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