For an ideal battery (r = 0 Ohm) closing the switch in the figure below does not

Question

For an ideal battery (r = 0 Ohm) closing the switch in the figure below does not affect the brightness of bulb A. In practice, bulb A dims just a little when the switch closes. To see why, assume that the 1.5 V battery has an internal resistance r = 0.42 Ohm and that the resistance of a glowing bulb is R = 5 Ohm. What is the current through bulb A when the switch is open? What is the current through bulb A after the switch has closed? By what percent does the current through A change when the switch is closed? (You must answer parts (a) and (b) before entering your answer.) Would closing the switch change the current through bulb A change if r = 0 Ohm? Yes No

Explanation / Answer

a) When switch is open,

current in circuit, I = e / (R + r) = 1.5 / (0.42 + 5)

I = 0.277 A

b) When switch is closed,

A and B are in parallel . so equivalent of these is R' then

1/R' = 1/R + 1/R

R' = R/2 = 5/2 = 2.5

current through battery, I = 1.5 / (2.5 + 0.42) = 0.514 A

current through bulb, i = I/2 = 0.257 A

c) %change = (change / initial current) x 100

= (0.277 - 0.257) x 100 / 0.277 = 7.22 %

d) No then it will not change.

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