Electricity — Free Physics Review Games.

Electric current is the rate of flow of electric charge, measured in amperes (A).

The ampere (A) is the SI unit of electric current, measuring the flow of charge per second.

A battery provides voltage (electrical potential difference) that drives current through a circuit.

There are two types of electric charge: positive (carried by protons) and negative (carried by electrons).

Ohm's Law states that voltage equals current times resistance: V = IR.

Using Ohm's Law: I = V/R = 12/4 = 3 A.

In series circuits, components share a single current path; in parallel circuits, current splits among multiple paths.

In series, total resistance is the sum of individual resistances: R_total = R1 + R2 + R3 + ...

Electrical power is the rate of energy transfer: P = IV, measured in watts (W).

Conductors are materials (like metals) with free electrons that allow electric current to flow easily.

1/R_total = 1/6 + 1/6 + 1/6 = 3/6 = 1/2, so R_total = 2 ohms.

In parallel, voltage is the same across each branch, but current divides among branches based on their resistance.

P = V^2/R, so R = V^2/P = 120^2/60 = 14400/60 = 240 ohms.

Kirchhoff's Current Law states that the sum of currents entering a junction equals the sum leaving, based on conservation of charge.

Energy = Power x Time = 100 W x 7200 s = 720,000 J (or 0.2 kWh).

The coulomb is the SI unit of electric charge, defined as the amount of charge transported by a current of one ampere in one second. An ampere, in contrast, is a unit of current, not charge itself, since current is the rate of charge flow. Students should remember that charge, current, voltage, and resistance each have distinct SI units that must not be interchanged.

Electrons carry a negative elementary charge of approximately \(-1.6\times10^{-19}\) C, which is fixed regardless of the atom they belong to. 'Positive' describes the charge of a proton, not an electron, so that choice confuses the two fundamental charge carriers. Recognizing that electrons are negative and protons are positive underlies all reasoning about current direction and charge transfer.

Like charges repel because the electric force between charges of the same sign is repulsive, as described by Coulomb's law. 'They attract each other' actually describes the interaction between opposite charges, not like charges. This attract/repel rule is the foundation for predicting the behavior of charged particles and objects in any electrostatics problem.

An insulator, such as rubber or glass, has tightly bound electrons that do not move freely, so it strongly resists the flow of charge through it. 'A material that allows electric charge to flow freely' instead describes a conductor, the opposite behavior. Knowing the distinction between conductors and insulators is essential for predicting how charge moves through different materials in a circuit.

Voltage, or potential difference, measures the energy transferred per unit charge as it moves between two points in a circuit. 'The rate at which charge flows through a wire' actually defines current, which is a separate quantity measured in amperes. Distinguishing voltage as an energy-per-charge quantity from current as a flow rate is key to correctly applying Ohm's law.

Resistance quantifies how strongly a material opposes the flow of current, and it is measured in ohms. 'The rate at which energy is delivered by a circuit' instead describes power, a different quantity calculated from voltage and current. Resistance is one of the three core quantities, along with voltage and current, linked together by Ohm's law.

Ohm's law states \(V = IR\), so dividing both sides by \(R\) gives \(I = \dfrac{V}{R}\), correctly isolating current. The choice '\(I = VR\)' incorrectly multiplies voltage and resistance rather than dividing, which would not have the right units for current. Being able to algebraically rearrange \(V=IR\) into any of its three forms is essential for solving circuit problems quickly.

An ammeter must be connected in series so that the same current flowing through the component also flows through the meter, allowing an accurate reading. Connecting it 'In parallel with the component being measured' would instead create a low-resistance shortcut that could damage the meter and alter the circuit's current. Remembering that ammeters go in series while voltmeters go in parallel is a key practical circuit-building rule.

A voltmeter is connected in parallel with a component so it measures the potential difference across that component without significantly altering the current flowing through it. 'In series with the component' would instead break the circuit path and add unwanted resistance, disrupting the very current being measured. This parallel placement rule for voltmeters contrasts directly with the series placement required for ammeters.

Current can only flow continuously when there is a complete, closed conducting loop connecting the power source back to itself through the circuit elements. 'An open switch at all times' would break this loop and prevent any current from flowing at all. This idea of a closed path is the starting point for analyzing any circuit diagram.

In a series circuit there is only one path for charge to flow, so by conservation of charge the current must be identical at every point around the loop. The claim that 'The current decreases after each resistor' is incorrect because charge is not consumed or lost as it passes through a resistor, only energy is dissipated. This constant-current property is a defining feature that distinguishes series circuits from parallel ones.

Each branch in a parallel circuit connects directly across the same two nodes, so each branch experiences the identical potential difference supplied by the source. 'The voltage decreases with each additional branch' confuses parallel circuits with series circuits, where voltage does divide among components. This equal-voltage property is what makes parallel circuits useful for household wiring, since each device receives full voltage.

Static electricity arises when charge builds up on an object's surface due to an imbalance between protons and electrons, often from friction between materials. 'The continuous flow of charge through a conductor' instead describes electric current, which is a moving charge phenomenon rather than a stationary buildup. Recognizing static charge buildup helps explain everyday phenomena like a spark from a doorknob after walking on carpet.

Using Ohm's law \(I = \dfrac{V}{R}\), dividing \(9\,V\) by \(3\,\Omega\) gives a current of \(3\,A\). The choice '\(27\,A\)' comes from mistakenly multiplying voltage and resistance instead of dividing them. Always identify which two quantities are given and rearrange \(V=IR\) to solve for the missing one.

By Ohm's law, resistance equals voltage divided by current, so \(R = \dfrac{20\,V}{2\,A} = 10\,\Omega\). The distractor '\(40\,\Omega\)' results from multiplying voltage and current rather than dividing, giving units of power instead of resistance. Keeping the units of each quantity in mind helps catch this type of calculation error.

In series, resistances simply add together, so the total resistance is \(5\,\Omega + 8\,\Omega = 13\,\Omega\). The value '\(3.08\,\Omega\)' comes from mistakenly applying the parallel resistance formula to resistors that are actually in series. Remembering that series resistances add directly while parallel resistances combine reciprocally prevents this common mix-up.

For resistors in parallel, \(\dfrac{1}{R_{eq}} = \dfrac{1}{6} + \dfrac{1}{3} = \dfrac{1}{2}\), so \(R_{eq} = 2\,\Omega\). The choice '\(9\,\Omega\)' simply adds the two resistances as though they were in series, which is not valid for a parallel combination. In parallel, the equivalent resistance is always smaller than the smallest individual resistor, a useful check on your answer.

Electrical power is calculated as \(P = IV\), so multiplying \(120\,V\) by \(2\,A\) gives \(240\,W\). The value '\(60\,W\)' results from dividing voltage by current instead of multiplying them, which does not correspond to the power formula. Power always combines both voltage and current, so both quantities must be multiplied together, not divided or added.

Squaring the current first gives \(I^2 = 4\,A^2\), and multiplying by resistance yields \(P = 4 \times 10\,\Omega = 40\,W\). The choice '\(20\,W\)' incorrectly uses \(I\) instead of \(I^2\), forgetting to square the current before multiplying by resistance. Because power depends on the square of current, doubling the current actually quadruples the dissipated power.

Squaring the voltage gives \(V^2 = 10000\,V^2\), and dividing by \(50\,\Omega\) gives \(P = 200\,W\). The choice '\(5000\,W\)' comes from forgetting to divide by resistance and only using part of the calculation incorrectly. This formula shows that power depends on the square of voltage, so small voltage changes can produce large power changes.

The total resistance is \(6\,\Omega\), giving a current of \(I = \dfrac{12\,V}{6\,\Omega} = 2\,A\), so the voltage across the \(4\,\Omega\) resistor is \(2\,A \times 4\,\Omega = 8\,V\). The distractor '\(4\,V\)' mistakenly assumes the two resistors split the voltage equally instead of proportionally to their resistance. In a series circuit, larger resistors always take a larger share of the total voltage.

Because both branches share the same voltage, the current splits inversely to resistance; solving for the equivalent resistance and voltage shows the \(2\,\Omega\) branch carries \(4\,A\) while the \(4\,\Omega\) branch carries \(2\,A\). The choice '\(3\,A\)' incorrectly assumes the current splits evenly regardless of resistance, ignoring that the smaller resistor always carries more current. Current divider problems require recognizing that lower-resistance paths draw proportionally more current.

Charge is the product of current and time, so \(Q = It = 4\,A \times 10\,s = 40\,C\). The value '\(2.5\,C\)' comes from dividing time by current instead of multiplying them, which does not match the definition of charge flow. This relationship \(Q=It\) connects the concepts of current and charge and is essential for problems involving charging times or capacitor charge.

Adding a resistor in parallel always lowers the equivalent resistance of the combination, and since \(I = \dfrac{V}{R}\), a lower resistance at constant voltage results in a higher total current drawn from the source. The claim that it 'decreases, since total resistance increases' misunderstands that parallel resistors decrease, not increase, the equivalent resistance. This is why adding more parallel appliances to a household circuit increases the total current the source must supply.

Since \(I = \dfrac{V}{R}\), current is inversely proportional to resistance at constant voltage, so doubling \(R\) cuts the current exactly in half. The choice 'It doubles' incorrectly treats current as directly proportional to resistance rather than inversely proportional. This inverse relationship is one of the most frequently tested consequences of Ohm's law.

Copper is a metal with loosely bound outer electrons that move freely, making it an excellent conductor of electric current. 'Rubber tubing' is a strong insulator whose electrons are tightly bound and do not move freely under an applied voltage. Choosing conductive materials like metals for wiring, versus insulating materials for coatings, is a practical application of conductor and insulator properties.

Because a series circuit has only one continuous path for current, breaking that path at any point stops current everywhere in the loop, so all bulbs go out. The choice 'They both get brighter' incorrectly assumes current can still flow, when in fact a broken filament creates an open circuit with no current at all. This vulnerability to a single point of failure is a key practical disadvantage of series wiring.

Because each bulb in a parallel circuit is connected across the same voltage through its own independent branch, removing one branch does not interrupt current flow through the others. The choice 'They immediately go out as well' would only be true in a series circuit, where all components share a single path. This independence of branches is why household wiring uses parallel circuits, so one broken appliance does not disable the entire circuit.

Converting power to kilowatts gives \(0.2\,kW\), and multiplying by the \(3\)-hour runtime gives an energy usage of $0.6\,kWh$. The choice '$600\,kWh$' results from forgetting to convert watts to kilowatts before multiplying by time. Energy in kilowatt-hours is simply power in kilowatts multiplied by time in hours, a calculation commonly used on electricity bills.

Electric charge is quantized because it is always transferred in discrete units equal to the charge of a single electron or proton, so any macroscopic charge is an integer multiple of this elementary charge. The claim that charge 'can be created or destroyed at any value' violates both quantization and the law of conservation of charge. Recognizing charge quantization explains why charge transfer, such as in charging by friction, always occurs in whole electron increments.

Series resistances simply add, so the total resistance is \(4\,\Omega + 6\,\Omega + 10\,\Omega = 20\,\Omega\). The value '\(1.76\,\Omega\)' incorrectly applies the parallel resistance formula, which would give a result smaller than the smallest resistor rather than larger. Series total resistance is always greater than or equal to the largest individual resistor in the chain.

Connecting resistors in parallel always produces an equivalent resistance smaller than either individual resistor, while connecting them in series always produces a sum larger than either resistor, so parallel resistance is always the smaller of the two configurations. The claim that they are 'always exactly equal' is only true in the special case where one resistor has zero or infinite resistance, which is not a general rule. This general comparison helps students quickly estimate whether a circuit's total resistance should be relatively small or large before doing detailed calculations.

The parallel combination of \(6\,\Omega\) and \(3\,\Omega\) reduces to \(\dfrac{6 \times 3}{6+3} = 2\,\Omega\), which then adds in series with the \(4\,\Omega\) resistor to give a total of \(6\,\Omega\). The choice '\(13\,\Omega\)' incorrectly adds all three resistors as though they were entirely in series, ignoring the parallel relationship between two of them. Solving combined circuits requires reducing each parallel or series section separately before combining the results.

The parallel section reduces to \(2\,\Omega\), giving a total circuit resistance of \(2\,\Omega + 2\,\Omega = 4\,\Omega\), so the total current, which equals the current through the series \(2\,\Omega\) resistor, is \(I = \dfrac{12\,V}{4\,\Omega} = 3\,A\). The choice '\(6\,A\)' comes from using only the series resistor's value in Ohm's law rather than the total circuit resistance. In a series-parallel circuit, the current through any purely series element equals the total current supplied by the source.

Kirchhoff's Voltage Law states that the sum of all voltage drops around a closed loop must equal the source voltage, so \(10\,V - 3\,V - 5\,V = 2\,V\) must be the remaining drop. The choice '\(8\,V\)' comes from simply adding the two known drops rather than subtracting them from the total source voltage. Applying Kirchhoff's Voltage Law lets you find unknown drops in any closed loop without knowing individual resistances.

Since \(P = \dfrac{V^2}{R}\) at fixed voltage, the parallel arrangement's lower equivalent resistance results in greater total power dissipation compared to the higher-resistance series arrangement. The claim that 'They dissipate exactly equal total power in both configurations' ignores that equivalent resistance differs significantly between series and parallel arrangements of the same resistors. This is why parallel circuits generally draw more current and dissipate more power from a fixed-voltage source than the same components wired in series.

Connecting all resistors in series produces the maximum possible total resistance because series resistances add directly, giving the largest combined opposition to current. The option 'All in parallel' instead minimizes total resistance, since parallel combinations always yield an equivalent resistance smaller than the smallest individual resistor. Knowing that series maximizes and parallel minimizes total resistance allows quick reasoning about circuit design without detailed calculation.

Adding the reciprocals gives \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6} = 1\), so the equivalent resistance is \(R_{eq} = 1\,\Omega\). The choice '\(11\,\Omega\)' mistakenly adds the three resistances directly, which is the series formula rather than the parallel one. Whenever combining unequal resistors in parallel, the reciprocal formula must be used rather than simple addition.

The voltage across a resistor in a divider is proportional to its share of total resistance, so \(V_9 = 12\,V \times \dfrac{9\,\Omega}{3\,\Omega+9\,\Omega} = 9\,V\). The choice '\(3\,V\)' incorrectly assigns the voltage proportional to the smaller resistor's fraction rather than the larger one. Voltage divider calculations always assign a larger voltage share to the resistor with greater resistance in the series chain.

Using the current divider rule, the current through one branch equals the total current times the other resistor divided by the sum, so \(I_6 = 10\,A \times \dfrac{4\,\Omega}{4\,\Omega+6\,\Omega} = 4\,A\). The choice '\(6\,A\)' incorrectly uses the \(6\,\Omega\) resistor's own value in the numerator rather than the opposite resistor's value. In a current divider, the branch with the larger resistance always carries the smaller share of current, opposite to how voltage dividers work.

The total resistance needed for a current of \(3\,A\) at \(24\,V\) is $R_{total} = \dfrac{24\,V}{3\,A} = 8\,\Omega$, and subtracting the known \(2\,\Omega\) resistor leaves \(R = 6\,\Omega\). The choice '\(8\,\Omega\)' mistakes the total circuit resistance for the value of the unknown resistor alone, forgetting to subtract the known resistor's contribution. Solving for an unknown component in a series circuit requires first finding the total resistance from Ohm's law, then isolating the unknown by subtraction.

Since \(P = \dfrac{V^2}{R}\), power depends on the square of voltage, so doubling the voltage increases the power by a factor of \(2^2 = 4\). The choice 'It increases by a factor of \(2\)' incorrectly assumes power scales linearly with voltage rather than quadratically. This squared relationship means that small increases in voltage can cause disproportionately large increases in power dissipation and heating.

The parallel section reduces to \(2\,\Omega\), giving a total resistance of \(3\,\Omega\) and a total current of \(\dfrac{10\,V}{3\,\Omega} \approx 3.33\,A\), so the voltage across the parallel section is \(3.33\,A \times 2\,\Omega \approx 6.67\,V\). The choice '\(5\,V\)' incorrectly assumes the voltage splits evenly between the series resistor and the parallel block regardless of their actual resistance values. Multi-step series-parallel problems require reducing the circuit to a single equivalent resistance before applying Ohm's law to find the total current and then individual voltage drops.

Kirchhoff's Current Law, based on conservation of charge, requires that the total current entering a node equal the total current leaving it, so \(5\,A + 3\,A = 8\,A\) must exit through the remaining wire. The choice '\(2\,A\)' incorrectly subtracts the two entering currents instead of adding them, which would violate charge conservation at the node. This node-based conservation principle applies to any junction with multiple current paths, regardless of circuit complexity.

The terminal voltage equals the EMF minus the voltage drop across the internal resistance, so $V_{terminal} = 12\,V - (2\,A)(1\,\Omega) = 10\,V$. The choice '\(12\,V\)' ignores the internal resistance entirely, treating the battery as ideal even though real batteries always lose some voltage internally when delivering current. Accounting for internal resistance explains why a battery's usable voltage drops as it supplies more current, especially under heavy load.