Electric Current - AP Physics 2

$R = 5 \text{ Ω}$. Apply $R = \frac{V}{I} = \frac{10}{2} = 5$ Ω.

$V = IR$. Voltage equals current times resistance.

Resistance. Instrument for measuring electrical resistance.

$P = I^2R$. Power dissipated as heat in resistive circuits.

Using an ammeter. Ammeter connected in series with circuit.

$I = 3 \text{ A}$. Apply $I = \frac{V}{R} = \frac{12}{4} = 3$ A.

Resistance. Instrument for measuring electrical resistance.

$P = \frac{V^2}{R}$. Alternative power formula using voltage and resistance.

$R_{total} = R_1 + R_2 + \dots$. Individual resistances add in series circuits.

Flow of electric charge. Rate of charge flow through a conductor.

Coulomb (C). Amount of electric charge, named after Coulomb.

$R = \frac{\rho L}{A}$. Resistance depends on material, length, and cross-sectional area.

Ohm ($\text{Ω}$). Named after Georg Ohm, symbol omega.

$P = IV$. Power equals current times voltage.

$I = \frac{q}{t}$. Current is charge divided by time.

$E = 200 \text{ Wh}$. Apply $E = Pt = 100 \times 2 = 200$ Wh.

Current increases. Direct proportional relationship from Ohm's Law.

$R_{total} = 3.33 \text{ Ω}$. Apply $\frac{1}{R_{total}} = \frac{1}{5} + \frac{1}{10} = \frac{3}{10}$, so $R_{total} = 3.33$ Ω.

Drift towards positive terminal. Electrons move opposite to conventional current direction.

Total resistance increases. Resistors in series add directly.

Ampere (A). Standard unit defined as one coulomb per second.

$P = 12 \text{ W}$. Apply $P = I^2R = 2^2 \times 3 = 12$ W.

$V = 9 \text{ V}$. Apply $V = IR = 1.5 \times 6 = 9$ V.

$V = \frac{W}{q}$. Voltage is work per unit charge.

$V = IR$. Voltage equals current times resistance.

Resistance increases. Higher temperature increases atomic vibrations.

Total resistance decreases. Parallel paths reduce overall resistance.

From positive to negative terminal. Conventional current flows opposite to electron flow.

$1 \text{ mA} = 0.001 \text{ A}$. Milli prefix means one thousandth.