Answer:
[tex]R=4.54\ \Omega[/tex]
Explanation:
It is given that,
Voltage of the battery, V = 1.5 V
Internal resistance of battery 1, [tex]r_1=0.32\ \Omega[/tex]
Internal resistance of battery 2, [tex]r_2=0.14\ \Omega[/tex]
Current flowing in the lamp, [tex]I=600\ mA=600\times 10^{-3}\ A=0.6\ A[/tex]
Total internal resistance of tow batteries,
[tex]r=r_1+r_2[/tex]
[tex]r=0.32+0.14[/tex]
[tex]r=0.46\ \Omega[/tex]
Let R is the resistance of the bulb. Let V is the total emf of the circuit. It is given by :
[tex]V=I(R+r)[/tex]
[tex]R=\dfrac{V}{I}-r[/tex]
[tex]R=\dfrac{2\times 1.5}{0.6}-0.46[/tex]
[tex]R=4.54\ \Omega[/tex]
So, the resistance of the bulb is 4.54 ohms. Hence, this is the required solution.
