A 12-V automobile storage battery has an internal resistance of 0.004 ohms. A) Find the current and power dissipation when the terminals are connected by a conductor of negligible Resistance B) The battery delivers 80A to the starter motor. What resistance would draw the same current? C) How much power is supplied to the motor?
i = V/R = 12 / .004 amps
P = i V = 144/R Joules
i = 80
R = V/i = 12/ 80
external resistance Rx = R - .004
power to motor = i^2 Rx = 6400 Rx
A) To find the current and power dissipation when the terminals are connected by a conductor of negligible resistance, we will need to use Ohm's Law, which states that V = I * R, where V is the voltage, I is the current, and R is the resistance.
Given:
Voltage (V) = 12V
Internal Resistance (r) = 0.004 ohms
Resistance of the conductor (R) = 0
Using Ohm's Law, we can calculate the current (I) flowing through the circuit:
V = I * (r + R)
12V = I * (0.004 ohms + 0 ohms)
12V = I * 0.004 ohms
I = 12V / 0.004 ohms
I = 3000A
So, the current flowing through the circuit is 3000A.
To calculate the power dissipation, we can use the formula P = I^2 * R, where P is the power dissipation, I is the current, and R is the resistance:
P = I^2 * R
P = (3000A)^2 * 0
P = 0W
Therefore, when the terminals are connected by a conductor of negligible resistance, the current flowing through the circuit is 3000A, and the power dissipation is 0W.
B) Now, let's calculate the resistance needed to draw the same current of 80A. We can rearrange Ohm's Law to solve for resistance:
V = I * R
R = V / I
R = 12V / 80A
R = 0.15 ohms
So, the resistance needed to draw the same current of 80A is 0.15 ohms.
C) To calculate the power supplied to the motor, we can use the formula P = I * V, where P is the power, I is the current, and V is the voltage:
P = I * V
P = 80A * 12V
P = 960W
Therefore, the power supplied to the motor is 960W.