For an ideal battery ( r = 0 ), closing the switch in (Figure 1) does not affect

Question

For an ideal battery (r = 0 ), closing the switch in (Figure 1) 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 1.50 V battery has an internal resistance r = 0.60 and that the resistance of a glowing bulb is R = 6.00 .

1- What is the current through bulb A when the switch is open?

2-What is the current through bulb A after the switch has closed?

3- By what percent does the current through A change when the switch is closed?

Explanation / Answer

two bulbs are in paralle connection so

1/Reff = 1/R1 + 1/R2

Reff = 6/2 = 3ohm

total resistance = 3+0.6 = 3.6 ohm

current i = V/Reff = 1.5/3.6 = 0.416 A

voltage across internal resistance = i*r = 0.416*0.6 = 0.2496 V

the voltage across the resistors is V = 1.5 - 0.2496 = 1.2504V

current through bulb = V/R = 1.2504/6 = 0.2084 A

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