a portable electric drill produces 1.2 hp at full load. If 85% of the power provided by the 9.6 V battery pack is useful, what is the current flow? how much power goes into the waste heat?
I don't know what to do. In class, we learned about amps, volts,and resistance, but never horsepower.
the equations we used were like P=v^2/R or V=iR and stuff like that.
please help
how to use formula in unit
To solve this problem, we can start by converting horsepower (hp) to watts (W) since the equations you mentioned involve watts:
1 horsepower (hp) is equal to approximately 746 watts (W).
Given that the drill produces 1.2 hp at full load, we can calculate the power produced by the drill in watts:
Power (W) = 1.2 hp * 746 W/hp
Power (W) = 895.2 W
Next, we need to find the current flow. We can use the equation:
Power (W) = Voltage (V) * Current (I)
Given that 85% of the power provided by the battery pack is useful, we can calculate the power provided by the battery pack:
Power (W) = 85% * Power produced (W)
Power (W) = 0.85 * 895.2 W
Power (W) = 760.92 W
We know the voltage of the battery pack is 9.6 V:
Power (W) = Voltage (V) * Current (I)
760.92 W = 9.6 V * Current (I)
Now we can solve for the current flow (I):
Current (I) = 760.92 W / 9.6 V
Current (I) ≈ 79.21 A (amperes)
Therefore, the current flowing is approximately 79.21 A.
To find the power that goes into waste heat, we can subtract the useful power from the total power:
Power into waste heat (W) = Total power (W) - Useful power (W)
Power into waste heat (W) = 895.2 W - 760.92 W
Power into waste heat (W) ≈ 134.28 W
Therefore, approximately 134.28 W goes into waste heat.
To solve this problem, we can use a combination of equations involving power, voltage, and current. Let's break it down step by step.
First, let's convert horsepower (hp) to watts (W) since we're dealing with electrical power. One horsepower is equal to approximately 746 watts, so we can calculate the power produced by the drill as follows:
Power (in watts) = 1.2 hp * 746 W/hp
Power (in watts) = 895.2 W
Next, we need to find the current flow. We can use Ohm's Law, which states that the current (I) flowing through a circuit is equal to the voltage (V) divided by the resistance (R). However, we don't have resistance information in this question. Instead, we are given the voltage and the efficiency of the battery pack.
The power provided by the battery pack is the useful power plus the wasted power:
Power (in watts) = Useful power (in watts) + Wasted power (in watts)
Given that 85% of the power provided by the battery pack is useful, we can express this relationship as:
Power (in watts) = 85% * Power provided by the battery pack
Substituting the given value of the total power (895.2 W), we can solve for the power provided by the battery pack:
Power provided by the battery pack = 895.2 W / 85%
Power provided by the battery pack = 1053.18 W
Now, let's use Ohm's Law to find the current flow. We can rearrange the formula V = IR to solve for I:
I = V / R
Since we don't have the resistance information, we can rearrange the power equation P = IV to solve for I:
I = P / V
Substituting the given values, we have:
I = 1053.18 W / 9.6 V
I ≈ 109.77 A
So the current flow is approximately 109.77 A.
Finally, to find the power that goes into waste heat, we can subtract the useful power from the total power provided by the battery pack:
Wasted power (in watts) = Total power provided by the battery pack - Useful power (in watts)
Wasted power (in watts) = 1053.18 W - 895.2 W
Wasted power (in watts) ≈ 157.98 W
Therefore, approximately 157.98 watts of power goes into waste heat.
I hope this explanation helps you understand the process behind solving this problem using the principles of electricity and power.
The produced power is useful power, and 1.2 hp is 895 W. (1 HP = 746 watts). Therefore the electrical input power is 895/0.85 = 1053 W.
Since electrical power P = I*V, the current is P/V = 1053/9.6 = 109.7 A
The waste heat is 15% of the input power, or 158 W