The lightbulb in the circuit diagram of (Figure 1) has a resistance of 0.80 . Co

Question

The lightbulb in the circuit diagram of (Figure 1) has a resistance of 0.80 . Consider the potential difference between pairs of points in the figure. Suppose that E = 4.0 V .

https://session.masteringphysics.com/problemAsset/1820041/6/Knight3e_EOC_ch23_p23.5.jpg

1- What is the magnitude of V12?

2- What is the magnitude of V23?

3-What is the magnitude of V34?

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

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

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

Explanation / Answer

apply loop in the circuit

E - 2 I - 0.80 I = 0

2.8 I = 4

I = 1.428 Amp

(a) so voltage

V12 = 1.428 x 2 = 2.857 volt Ans

(b)   

V23 = 1.428 x 0.80 = 1.1424 volt Ans

(c) there is no resistance

V34 = 1.428 x zero = 0 volt Ans

when bulb is removed the the current

I = 4 / 2 = 2 Amp

(d)

so the voltage V12 = 2 x 2 = 4 volt Ans

(e) there is no resistance

V23 = 2 x zero = zero volt = 0 volt Ans

(f) there is no resistance

V34 = 2 x zero = 0 volt Ans

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