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 0.320 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?

Answer 1

Answer:

$R=4.54\ \Omega$

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.