Physics, 15.11.2019 18:31 ValeryGi9733

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 0.320 ω, the other an internal resistance of 0.140 ω. when the switch is closed, a current of 600 ma occurs in the lamp. (a) what is the bulb's resistance?

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Answer from: dward5823

$R=4.54\ \Omega$

Explanation:

It is given that,

Voltage of the battery, V = 1.5 V

Internal resistance of battery 1, $r_1=0.32\ \Omega$

Internal resistance of battery 2, $r_2=0.14\ \Omega$

Current flowing in the lamp, $I=600\ mA=600\times 10^{-3}\ A=0.6\ A$

Total internal resistance of tow batteries,

r=r_1+r_2

r=0.32+0.14

$r=0.46\ \Omega$

Let R is the resistance of the bulb. Let V is the total emf of the circuit. It is given by :

V=I(R+r)

$R=\dfrac{V}{I}-r$

$R=\dfrac{2\times 1.5}{0.6}-0.46$

$R=4.54\ \Omega$

So, the resistance of the bulb is 4.54 ohms. Hence, this is the required solution.

Answer from: Quest

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Answer from: Quest

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