Asked by homie

Batteries consist of an ideal source of potential difference in series with a small resistance. The electrical energy of the battery is produced by chemical reactions that occur in the battery. However, these reactions also result in a small resistance that, unfortunately, cannot be completely eliminated. A flashlight contains two batteries in series. Each has a potential difference of 1.50 V and an internal resistance of 0.18 . The bulb has a resistance of 29.0 .

(a) What is the current through the bulb?
(b) How much power does the bulb dissipate?
(c) How much greater would the power be if the batteries had no internal resistance?

last question, and I do not even know where I went wrong, help I tried and nothing is working
Answered by Anonymous
total R = 2 * 0.18 + 29 = 29.36 ohms
i = V/R = 2 * 1.50 / 29.36= 0.102 amps answer (a) but through everything
(b) bulb power = i^2 R = 0.102^2 * 29 = 0.303 watts (do not use V^2/R because the whole V is not on the bulb)
(c) new i = V/R = 2*1.5 / 29 = 0.103 amps
so
bulb power = i^2 R = 0.103^2 * 29 = 0.310 watts
.310 - .303 = 0.007 watts more
Answered by homie
God bless you, Anonymous.
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