Two 1.50-V batteries—with their positive terminals in the same direction—are inserted in series into the barrel of a flashlight. One battery has an internal resistance of 0.27 Ω, the other an internal resistance of 0.11 Ω. When the switch is closed, a current of 600 mA occurs in the lamp.What is the bulb's resistance?Answer for part 1What fraction of the chemical energy transformed appears as internal energy in the batteries? Write the answer as a % (ie out of 100)
Question
Two 1.50-V batteries—with their positive terminals in the same direction—are inserted in series into the barrel of a flashlight. One battery has an internal resistance of 0.27 Ω, the other an internal resistance of 0.11 Ω. When the switch is closed, a current of 600 mA occurs in the lamp.What is the bulb's resistance?Answer for part 1What fraction of the chemical energy transformed appears as internal energy in the batteries? Write the answer as a % (ie out of 100)
Solution
To find the bulb's resistance, we first need to find the total voltage and total resistance in the circuit.
The total voltage (V_total) is the sum of the voltages of the two batteries, which are connected in series. So, V_total = 1.50V + 1.50V = 3.00V.
The total resistance (R_total) is the sum of the internal resistances of the two batteries and the resistance of the bulb (R_bulb). So, R_total = R_bulb + 0.27Ω + 0.11Ω.
We know that the current (I) in the circuit is 600mA or 0.6A. According to Ohm's law (V = IR), we can find the resistance of the bulb:
R_bulb = V_total / I - (0.27Ω + 0.11Ω) = 3.00V / 0.6A - 0.38Ω = 5Ω - 0.38Ω = 4.62Ω.
For the second part of the question, we need to find the power dissipated in the batteries (P_batteries) and the total power supplied by the batteries (P_total).
The power dissipated in the batteries is the sum of the power dissipated in each battery, which can be found using the formula P = I^2R. So, P_batteries = I^2 * (0.27Ω + 0.11Ω) = 0.6A^2 * 0.38Ω = 0.1368W.
The total power supplied by the batteries is the product of the total voltage and the current, P_total = V_total * I = 3.00V * 0.6A = 1.8W.
The fraction of the chemical energy transformed into internal energy in the batteries is then P_batteries / P_total = 0.1368W / 1.8W = 0.076 or 7.6% when expressed as a percentage.
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