Answer:
0.95 A
Explanation:
P = Power of bulb = 4 W
V = Voltage = 12 V
Power is given by
[tex]P=\dfrac{V^2}{R}\\\Rightarrow R=\dfrac{V^2}{P}\\\Rightarrow R=\dfrac{12^2}{4}\\\Rightarrow R=36\ \Omega[/tex]
The connections are in Parallel
Equivalent resistance of these two bulb
[tex]\dfrac{36}{2}=18\ \Omega[/tex]
Current from the source
[tex]I=\dfrac{11.4}{18}=0.63\ A[/tex]
The voltage drop is
[tex]12-11.4=0.6\ V[/tex]
Voltage drop is given by
[tex]\Delta V=R_iI\\\Rightarrow R_i=\dfrac{\Delta V}{I}\\\Rightarrow R_i=\dfrac{0.6}{0.63}\\\Rightarrow R_i=0.952380952381\ A[/tex]
The internal resistance is 0.95 A
