Answer to Question #280870 in Physics for azzam

Question

An automobile battery has an emf of 12.6 V and an internal resistance of 0.080 Ω. The headlights together have an equivalent resistance of 5.00 Ω (assumed constant). What is the potential difference across the headlight bulbs (a) when they are the only load on the battery and (b) when the starter motor is operated, with 35.0 A of current in the motor?

Answer 1

(a) In this case we can write the emf as follows:

$$\Epsilon=I(R+r),$$ $$I=\dfrac{E}{R+r}=\dfrac{12.6\ V}{5\ \Omega+0.08\ \Omega}=2.48\ A.$$

Finally, we can find the potential difference across the headlight bulbs from the Ohm's law:

$$V=IR=2.48\ A\times5\ \Omega=12.4\ V.$$

(b) Let's $I_1$ is the current that flows through the automobile battery, $I_2$ is the current that flows through the headlight bulbs. Then, from the Kirchhoff's law, we get:

$$I_1=I_2+35\ A.$$

We can find the emf as follows:

$$\Epsilon=I_1r+I_2R=(I_2+35\ A)r+I_2R.$$

From this formula we can find $I_2$:

$$I_2=\dfrac{E-35\ A\times r}{R+r}=\dfrac{12.6\ V-35\ A\times0.08\ \Omega}{5\ \Omega+0.08\ \Omega}=1.93\ A.$$

Finally, we can find the potential difference across the headlight bulbs from the Ohm's law:

$$V=I_2R=1.93\ A\times5\ \Omega=9.65\ V.$$