The lightbulb in the circuit diagram of (Figure 1) has a resistance of 1.1 . Con

Question

The lightbulb in the circuit diagram of (Figure 1) has a resistance of 1.1 . Consider the potential difference between pairs of points in the figure. Suppose thatE = 3.5 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

Since bulb and resistor are in series

Rnet = 2+1.1=3.1ohm

Current in the circuit = I = V/R = 1.129A

Using V=IR

A) V=2*1.129=2.258V

B) V = 1.1*1.129 = 1.2419V

C) V= 0*1.129 = 0V (NEGLECTING RESISTANCE OF CONNECTING WIRE)

D) V= 0V, (NO CURRENT IN THE CIRCUIT SINCE IT IS NOT CLOSED)

E) V= 0V, (NO CURRENT IN THE CIRCUIT SINCE IT IS NOT CLOSED)

F) V= 0V, (NO CURRENT IN THE CIRCUIT SINCE IT IS NOT CLOSED)

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