For an ideal battery (r = 0 ohms), closing the switch in the figure below does n

Question

For an ideal battery (r = 0 ohms), 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 2.92 V battery has an internal resistance r = 1.29 ohms and that the resistance of a glowing bulb is R = 6.00 ohms.

a) What is the current through bulb A when the switch is open?

b) What is the current through bulb A after the switch has closed?

c) By what percentage does the current through A change when the switch is closed?

Explanation / Answer

a) current through bulb A, IA = e/Rnet

= e/(r + RA)

= 2.92/(1.29 + 6)

= 0.4005 A

b) when switch is closed

net resitance of the ckt, Rnet = r + RA*RB/(RA+Rb)

= 1.29 + 6*6/(6+6)

= 4.29 ohms

current through battery, I = e/Rnet

= 2.92/4.29

= 0.68 A

current through A, IA = I/2

= 0.68/2

= 0.34 A

c)

IA_closed/I_open = 0.34/0.4005

= 0.849

IA_closed = I_open*0.849

= I_open*(1 - 0.151)

IA_closed = I_open - 15.1% of I_open

so, current decreases by 15.1%

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