An automobile battery has an emf of 12.6 V and an internal resistance of 0.0500 . The headlights together present equivalent resistance 5.60 (assumed constant).

(a) What is the potential difference across the headlight bulbs when they are the only load on the battery?
V
(b) What is the potential difference across the headlight bulbs when the starter motor is operated, taking an additional 35.0 A from the battery?

a. I=E/(Ri+Rb) = (12.6/(0.05+5.60)=2.23A

V = I*Rb = 2.23 * 5.60 = 12.49 Volts.

b. Vb=E-I*Ri=12.6 - (35+2.23)0.05=10.74
Volts.

(a) Why did the headlight bulbs cross the road? To find the potential difference!

In this case, when the headlights are the only load on the battery, we can find the potential difference using Ohm's Law:

V = E - I * r

where V is the potential difference, E is the emf of the battery, I is the current, and r is the internal resistance.

Using the given values, we have:

V = 12.6 V - 0.0500 Ω * 5.60 A

Calculating this, we find:

V ≈ 12.6 V - 0.280 V

V ≈ 12.32 V

So, the potential difference across the headlight bulbs when they are the only load on the battery is approximately 12.32 V.

(b) Now, when the starter motor is operated and an additional 35.0 A is taken from the battery, we can use the same formula to find the potential difference.

V = E - I * r

V = 12.6 V - (5.60 A + 35.0 A) * 0.0500 Ω

Calculating this, we get:

V ≈ 12.6 V - 2.30 V

V ≈ 10.3 V

So, the potential difference across the headlight bulbs when the starter motor is operated, taking an additional 35.0 A from the battery, is approximately 10.3 V.

To solve these questions, we will use Ohm's Law and Kirchhoff's Voltage Law (KVL).

(a) When the headlights are the only load on the battery, the equivalent resistance is 5.60 Ω.

Step 1: Calculate the current flowing through the circuit using Ohm's Law:

I = V / R
Where I is the current, V is the emf of the battery, and R is the equivalent resistance.

Plugging in the values:
I = 12.6 V / 5.60 Ω
I ≈ 2.25 A

Step 2: Calculate the potential difference across the headlight bulbs using Ohm's Law:

V = I * R
Where V is the potential difference, I is the current, and R is the equivalent resistance.

Plugging in the values:
V = 2.25 A * 5.60 Ω
V ≈ 12.6 V

(b) When the starter motor is operated, an additional 35.0 A is drawn from the battery.

Step 1: Calculate the total current flowing through the circuit:

I_total = I_headlights + I_starter_motor
Where I_total is the total current, I_headlights is the current flowing through the headlights (calculated in part (a)), and I_starter_motor is the additional current drawn.

Plugging in the values:
I_total = 2.25 A + 35.0 A
I_total ≈ 37.25 A

Step 2: Calculate the potential difference across the headlight bulbs using Ohm's Law:

V = I * R
Where V is the potential difference, I is the current, and R is the equivalent resistance.

Plugging in the values:
V = 37.25 A * 5.60 Ω
V ≈ 208.55 V

So, the potential difference across the headlight bulbs when they are the only load on the battery is approximately 12.6 V, and when the starter motor is operated, it is approximately 208.55 V.

To answer these questions, we need to use Ohm's Law and the concept of series circuit.

First, let's calculate the current flowing through the circuit in both cases.

(a) When the headlights are the only load on the battery:
Using Ohm's Law, we can calculate the current (I) flowing through the circuit by dividing the emf (E) by the total resistance (R), which is the sum of the internal resistance (r) of the battery and the resistance of the headlights (R_headlights).
So, the current through the circuit is given by:
I = E / (r + R_headlights)

Given:
Emf of the battery, E = 12.6 V
Internal resistance of the battery, r = 0.0500 Ω
Resistance of the headlights, R_headlights = 5.60 Ω

Plugging in the values, we get:
I = 12.6 V / (0.0500 Ω + 5.60 Ω)

Calculating this, you will get the value of current flowing through the circuit.

(b) When the starter motor is operated, taking an additional 35.0 A from the battery:
In this case, we need to consider the total current flowing through the circuit, which is the sum of the current through the headlights (calculated in part a) and the additional current taken by the starter motor.

Given:
Additional current taken by the starter motor, I_additional = 35.0 A

So, the total current flowing through the circuit is given by:
I_total = I_headlights + I_additional

Now, we can calculate the potential difference across the headlight bulbs using Ohm's Law. The potential difference (V) across the headlights is given by the product of current (I_headlights) flowing through them and their resistance (R_headlights).
So,
V = I_headlights * R_headlights

Calculating this using the values obtained in part a, you will get the potential difference across the headlight bulbs when the starter motor is operated.

Make sure to substitute the correct values and units into the equations to find the final answers.

(from chapter 12) At a temperature of 0 °C, the mass and volume of a fluid are 825 kg and 1.17 m3. The coefficient of volume expansion is 1.26 × 10 – 3(C ° ) – 1. (a) What is the density of the fluid at this temperature? (b) What is the density of the fluid when the temperature has risen to 20.0 °C?

2.