Electric Current & Circuits Explained, Ohm's Law, Charge, Power, Physics Problems, Basic Electricity

The video introduces conventional current flow from the positive to the negative terminal of a battery and contrasts it with electron flow. It defines electric current as the rate of charge flow (Q/T), measured in amperes (amps), where 1 amp equals 1 Coulomb per second. Ohm's Law (V=IR) is then introduced, explaining the relationship between voltage, current, and resistance. It discusses how voltage and current are directly related, while resistance and current are inversely related, using a highway analogy.

Electric power is defined as the product of voltage and current (P=VI), measured in watts (W), where 1 watt equals 1 Joule per second. The video also presents two alternative forms of the power equation derived from Ohm's Law: P = I²R and P = V²/R.

This problem involves a current of 3.8 amps flowing for 12 minutes. The first part calculates the total charge passed (Q = I*T) by converting minutes to seconds, resulting in 2736 Coulombs. The second part determines the number of electrons this charge represents, given that one electron has a charge of 1.6 x 10^-19 Coulombs, leading to approximately 1.71 x 10^22 electrons.

This problem examines a 9V battery connected to a 250-ohm resistor. Using Ohm's Law (I=V/R), the current through the resistor is calculated as 0.036 amps (or 36 milliamps). The power dissipated by the resistor is then found using P = I²R, resulting in 0.324 watts (or 324 milliwatts). The problem also verifies that the power delivered by the battery (P=VI) is the same as the power dissipated by the resistor.

A 12-volt battery powers a light bulb drawing 150 milliamps. First, the electrical resistance of the bulb is calculated using Ohm's Law (R=V/I) after converting milliamps to amps, yielding 80 ohms. Next, the power consumed by the bulb is determined using P=VI, found to be 1.8 watts. Finally, the cost to operate the bulb for a month (30 days, 24 hours/day) is calculated, converting power to kilowatts and multiplying by the electricity cost of 11 cents per kilowatt-hour, resulting in approximately 14 cents.

This problem deals with a motor using 50 watts of power and drawing 400 milliamps. The voltage across the motor is found using P=VI, resulting in 125 volts after converting milliamps to amps. The internal resistance of the motor is then calculated using Ohm's Law (R=V/I), which comes out to 312.5 ohms.

12.5 Coulombs of charge flow through a 5 kilo-ohm resistor in 8 minutes. The electric current is calculated first (I=Q/T) by converting minutes to seconds, yielding 0.026 amps (or 26 milliamps). The power consumed by the resistor is then calculated using P = I²R, converting kilo-ohms to ohms, resulting in 3.38 watts. Lastly, the voltage across the resistor is determined using Ohm's Law (V=IR), which is 130 volts.