Answer:
(a) [tex]V=11.86\ V[/tex]
(b) [tex]V=9.76\ V[/tex]
Explanation:
Electric Circuits
Suppose we have a resistive-only electric circuit. The relation between the current I and the voltage V in a resistance R is given by the Ohm's law:
[tex]V=R.I[/tex]
(a) The electromagnetic force of the battery is [tex]\varepsilon =12.6\ V[/tex] and its internal resistance is [tex]R_i=0.06\ \Omega[/tex]. Knowing the equivalent resistance of the headlights is [tex]R_e=5.2\ \Omega[/tex], we can compute the current of the circuit by using the Kirchhoffs Voltage Law or KVL:
[tex]\varepsilon=i.R_i+i.R_e=i.(R_i+R_e)[/tex]
Solving for i
[tex]\displaystyle i=\frac{\varepsilon}{ R_i+R_e}=\frac{12}{0.06+5.2}=2.28\ A[/tex]
i=2.28\ A
The potential difference across the headlight bulbs is
[tex]V=\varepsilon -i.R_i=12\ V-2.28\ A\cdot 0.06\ \Omega=11.86\ V[/tex]
[tex]V=11.86\ V[/tex]
(b) If the starter motor is operated, taking an additional 35 Amp from the battery, then the total load current is 2.28 A + 35 A = 37.28 A. Thus the output voltage of the battery, that is the voltage that the bulbs have is
[tex]V=\varepsilon -i.R_i=12\ V-37.28\ A\cdot 0.06\ \Omega=9.76\ V[/tex]
