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

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 2.67 V battery has an internal resistance r = 1.82 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

Here,

a)

when the switch is open

Using Ohm's Law

Current in Bulb A , IAi = E/(r + R)

IAi = 2.67/(1.82 + 6)

IAi = 0.341 A

the current through bulb A when the switch is open is 0.341 A

b)

when the switch is closed

for the equivalent resistance

Req = r + RA * RB/(RA + RB)

Req = 1.82 + 6 * 6/(6 + 6)

Req = 4.82 Ohm

current in A , IAf = (2.68/4.82) * (6/(6 + 6))

IAf = 0.278 A

the current through bulb A after the switch has closed is 0.278 A

c)

percentage of current in A = (IAf - IAi)/IAi * 100

percentage of current in A = (0.278 - 0.341)/0.341 * 100

percentage of current in A = -18.75 %

percentage does the current through A change when the switch is closed is -18.75 %

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