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

Question

For an ideal battery (r = 0 ?) 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 = 0.52 ? and that the resistance of a glowing bulb is R = 7 ?.

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

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

(c) 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.)
%
(d) Would closing the switch change the current through bulb A change if r = 0 ??

YesNo   

For an ideal battery (r = 0 ?) 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 = 0.52 ? and that the resistance of a glowing bulb is R = 7 ?. (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 percent does the current through A change when the switch is closed? (You must answer parts (a) and (b) before entering your answer.) (d) Would closing the switch change the current through bulb A change if r = 0 ??

Explanation / Answer

I see you TTU student.

A) Since V = IR, I = V/Req

Req = r + R = 0.52 + 7 = 7.52

I = 1.5/7.52 = 0.199 or 199mA

B) Req = r + R//R = 0.52 + 1/((1/7)+(1/7)) = 4.02 Ohms

I = 1.5/4.02 = 0.373A or 373mA. Divide the current by 2 since the current splits evenly along the 2 parallel lightbulb paths.

I = 0.1865A or 186.5mA

C) ((0.1865-0.199)/0.199) * 100 = 6.3%

D) No

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