Physics, 05.02.2020 12:43 JimmySample7

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 0.240 ω, the other an internal resistance of 0.180 ω. when the switch is closed, a current of 600 ma occurs in the lamp. (a) what is the bulb's resistance? ω (b) what fraction of the chemical energy transformed appears as internal energy in the batteries? %

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Answer from: NightCaribou644

a. The resistance of the bulb is equal to 4.58 Ohms.

b. The fraction of the chemical energy transformed which appears as internal energy in the batteries is 8.4%.

Given the following data:

Voltage of battery = 1.5 VoltsInternal resistance A = 0.240 OhmsInternal resistance B = 0.180 OhmsCurrent = 600 mA = 0.6 Amps

a. To find the resistance of the bulb:

First of all, we would determine the total electromotive force (E) of the electric circuit:

$E = 2 \times 1.50$

E = 3.0 Volts

Total internal resistance = $0.240 + 0.180 = 0.42 \;Ohms$

Mathematically, electromotive force (E) is given by the formula:

E = V + Ir ....equation 1.

Where:

E is the electromotive force (E).V is the the voltage or potential difference.I is the current.r is the internal resistance.

According to Ohm's law, voltage is given by:

V = IR ....equation 2

Substituting eqn 2 into eqn 1, we have:

$E = IR + Ir\\\\E = I(R + r)\\\\R + r = \frac{E}{I} \\\\R = \frac{E}{I} - r\\\\R = \frac{3}{0.6} - 0.42\\\\R = 5 - 0.42$

Resistance, R = 4.58 Ohms

b. To determine what fraction of the chemical energy transformed appears as internal energy in the batteries:

First of all, we would determine the electromotive force (E) in the batteries.

$E_B = IR\\\\E_B = 0.6 \times 0.42$

E_B = 0.252 Volt

$Percent = \frac{E_B}{E} \times 100\\\\Percent = \frac{0.252}{3} \times 100\\\\Percent = \frac{25.2}{3}$

Percent = 8.4 %

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Two 1.50-v batteries—with their positive terminals in the same direction—are inserted in series into...