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

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 thatE = 3.0V .

Part A. What is the magnitude of ?V12?

Part B. What is the magnitude of ?V23?

Part C.What is the magnitude of ?V34?

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 thatE = 3.0V . Part A. What is the magnitude of ?V12? Part B. What is the magnitude of ?V23? Part C.What is the magnitude of ?V34? What is the magnitude of ?V12if the bulb is removed from the socket (i.e. the circuit is not closed)?

Explanation / Answer

Part A:

2 ohm and 0.60 ohm resistances in series across a 3 V source

so R=2+0.60

=2.6ohm

The current I = 3 V/2.6 ohms = 1.153 A

DeltaV12 = voltage across the 2 ohm resistor

= IR

=1.153*2

DeltaV12 =2.30V

b)DeltaV23 = voltage across the 0.60 ohm resistor

= IR

=1.153*0.60

DeltaV23 = 0.6918 V

c)3 and 4 are the same (a wire connection between the two) so there is never a voltage between them.

so DeltaV34 = 0 V

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