7.) Two 1.50 V batteries -- with their positive terminals inthe same direction -

Question

7.) Two 1.50 V batteries -- with their positive terminals inthe same direction -- are inserted in series into the barrel of aflashlight. One battery has an internal resistance of 0.240 , the other an internal resistance of0.110 . When the switch is closed,a current of 600 mA occurs in the lamp. (a) What is the lamp's resistance?
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(b) What fraction of the chemical energytransformed appears as internal energy in the batteries?
___________________ % (a) What is the lamp's resistance?
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(b) What fraction of the chemical energytransformed appears as internal energy in the batteries?
___________________ %

Explanation / Answer

(a) total voltage = total resistance *current .        1.50 + 1.50 = (0.240 + 0.110 + R) * 0.600 .         3.00 = (0.250 + R ) * 0.600 .         R = 4.75 ohms   is the resistance of the bulb . (b) the total power supplied is   VI = 3.00 * 0.600 = 1.80 Watts . and transformed by the internal resistance is  I2R = 0.6002 * 0.25  =    0.090 W . So the percentage is    0.090 / 1.80 =    0.05 =   5.00%

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