2.4 - Electric Circuits

Electric current is the rate at which charge passes a point: $I=\Delta Q/\Delta t$. One ampere is one coulomb per second. Rearranging gives $\Delta Q=I\Delta t$.

Current can be carried by different charged particles. In a metal the mobile carriers are electrons, but conventional current is defined in the direction positive charge would move, opposite to electron drift.

A steady current of $0.40\,\mathrm{A}$ flows for $3.0\,\mathrm{min}=180\,\mathrm{s}$. The charge transferred is $Q=(0.40)(180)=72\,\mathrm{C}$. The number of electrons is $Q/e=4.5\times10^{20}$.

The potential difference between two points is the energy transferred per unit charge moving between them: $V=W/Q$. One volt is one joule per coulomb. Rearrangements are $W=VQ$ and $Q=W/V$.

Across a component, electrical energy is transferred to other stores. A larger p.d. means more energy is transferred for each coulomb, not necessarily that more charge flows; current also depends on the circuit resistance.

Moving $2.5\,\mathrm{C}$ through a p.d. of $12\,\mathrm{V}$ transfers $W=(12)(2.5)=30\,\mathrm{J}$. An electron accelerated through $108\,\mathrm{V}$ gains energy of magnitude $eV$.

Resistance at an operating point is defined by $R=V/I$ and measured in ohms. It compares the p.d. across a component with the current through it.

Ohm's law states that $I\propto V$ for a conductor at constant temperature. The ratio $V/I$ is then constant, so an $I$-against-$V$ graph is a straight line through the origin.

If $V=6.0\,\mathrm{V}$ produces $I=0.25\,\mathrm{A}$, then $R=6.0/0.25=24\,\Omega$. Doubling $V$ doubles $I$ only if conditions, especially temperature, keep $R$ constant.

Current division is a charge-flow statement; p.d. division is an energy-per-charge statement. A larger series resistance receives a larger share of p.d. because the same current flows and $V=IR$.

In series, current $I$ is common and p.d.s add. Using $V_T=V_1+V_2+\cdots$ and $V=IR$ gives $IR_T=IR_1+IR_2+\cdots$, so $R_T=R_1+R_2+\cdots$.

In parallel, p.d. $V$ is common and branch currents add. Using $I_T=I_1+I_2+\cdots$ and $I=V/R$ gives $V/R_T=V/R_1+V/R_2+\cdots$, so $1/R_T=1/R_1+1/R_2+\cdots$.

Two resistors $6.0\,\Omega$ and $3.0\,\Omega$ give $9.0\,\Omega$ in series but $2.0\,\Omega$ in parallel. A parallel equivalent must be smaller than the smallest branch resistance because it provides more paths for charge.

Substituting $V=IR$ into $P=VI$ gives $P=I^2R$. Substituting $I=V/R$ gives $P=V^2/R$. The choice matters when comparing circuits: at fixed current, power rises with $R$; at fixed p.d., power falls with $R$.

A $12\,\Omega$ resistor carries $0.50\,\mathrm{A}$. It dissipates $P=(0.50)^2(12)=3.0\,\mathrm{W}$ and transfers $W=(3.0)(40)=120\,\mathrm{J}$ in $40\,\mathrm{s}$.

At any point, resistance is $V/I$, not simply the gradient of an $I$-against-$V$ curve. A steeper ray from the origin corresponds to larger $I/V$ and therefore smaller resistance.

To obtain a graph, place an ammeter in series and a voltmeter across the component, vary p.d. safely, and reverse polarity when negative values are required. Include a current-limiting resistor for a diode.

For a uniform conductor, $R=\rho l/A$, where $l$ is length, $A$ is cross-sectional area and $\rho$ is resistivity in $\Omega\,\mathrm{m}$. Rearranging gives $\rho=RA/l$.

For the same material and temperature, resistance is directly proportional to length and inversely proportional to area. Doubling length doubles $R$; doubling diameter makes area four times larger and reduces $R$ to one quarter.

A wire has $R=3.0\,\Omega$, $l=2.0\,\mathrm{m}$ and $A=1.5\times10^{-6}\,\mathrm{m^2}$. Then $\rho=(3.0)(1.5\times10^{-6})/2.0=2.3\times10^{-6}\,\Omega\,\mathrm{m}$.

Measure wire diameter with a micrometer at several positions and orientations, correct any zero error, average it and calculate $A=\pi d^2/4$. Measure the test length $l$ between electrical contacts with a metre rule.

Connect the test wire and ammeter in series with a d.c. supply, switch and variable resistor; connect a voltmeter across the measured length. Use a low current and close the switch only for readings so heating does not change resistivity.

For several lengths, record $V$ and $I$ and calculate $R=V/I$. Plot $R$ vertically against $l$ horizontally. From $R=(\rho/A)l$, best-fit gradient $m=\rho/A$, so $\rho=mA$. A single-length result may instead use $\rho=RA/l$, but repeated lengths give a stronger test.

The current carried through cross-sectional area $A$ is $I=nqvA$. Here $n$ is mobile charge-carrier number density, $q$ is charge per carrier and $v$ is mean drift speed.

For a given area, current is larger when more mobile carriers are available or when their drift response is greater. Metals have a large density of conduction electrons and therefore conduct readily. Insulators have extremely few mobile carriers, producing tiny current for an applied field; semiconductors can change carrier number strongly with temperature or illumination.

If two wires carry the same current with the same $n$ and $q$, but one has four times the area, its drift speed is one quarter as large. Conversely, simultaneous fourfold increases in $I$ and $A$ leave $v$ unchanged.

A uniform wire has constant resistivity and cross-sectional area, so resistance from one end to distance $x$ is $R_x=\rho x/A$. In steady current $I$, the p.d. across that length is $V_x=IR_x$, hence $V_x\propto x$.

A graph of potential against distance is a straight line for a uniform wire carrying constant current. Its sign of gradient depends on the chosen direction and reference potential; its magnitude is the potential gradient.

If a wire drops $6.0\,\mathrm{V}$ uniformly over $1.5\,\mathrm{m}$, the drop is $4.0\,\mathrm{V\,m^{-1}}$. A point $0.40\,\mathrm{m}$ from the high-potential end is $1.6\,\mathrm{V}$ below that end.

Two resistors $R_1$ and $R_2$ in series carry the same current $I=V_{in}/(R_1+R_2)$. The output across $R_2$ is therefore $V_{out}=IR_2=V_{in}R_2/(R_1+R_2)$.

With $V_{in}=12\,\mathrm{V}$, $R_1=2.0\,\mathrm{k\Omega}$ and $R_2=4.0\,\mathrm{k\Omega}$, the output across $R_2$ is $12(4.0/6.0)=8.0\,\mathrm{V}$.

For an unloaded divider, the output across a component is $V_{out}=V_{in}R_{out}/R_{total}$. First mark exactly which resistor the output spans; then change the sensor resistance and recompute the ratio.

A control circuit switches when $V_{out}$ crosses a threshold. To predict its response, find sensor resistance at the stated condition, calculate output, then determine how the output moves on either side of that condition.

When a cell supplies current, energy conservation gives $\varepsilon=V+Ir$, so $V=\varepsilon-Ir$. Increasing current increases the lost volts and lowers terminal p.d. if $\varepsilon$ and $r$ remain constant.

With negligible current, $Ir\approx0$ and terminal p.d. is approximately the e.m.f. Under load, the source behaves like an ideal e.m.f. in series with its internal resistance.

Connect the cell, ammeter, switch and variable resistor in series. Connect a voltmeter directly across the cell terminals. Use high-resistance voltmeter and vary the external resistance to obtain a safe range of current values.

For each setting, close the switch briefly, record current $I$ and terminal p.d. $V$, then open it to reduce heating and cell discharge. Repeat readings and do not include a zero-resistance short circuit.

From $V=\varepsilon-Ir$, plot $V$ vertically against $I$ horizontally. Draw a best-fit line: the vertical intercept is $\varepsilon$, and the negative gradient is $-r$, so internal resistance is the positive magnitude of the gradient.

Use a large gradient triangle, include $\mathrm{V\,A^{-1}}=\Omega$, and inspect scatter or curvature before accepting a constant-$r$ model. The intercept is extrapolated, so a well-spread data range improves it.

For the same applied p.d., a metal's rising resistance reduces current. In an NTC thermistor, higher temperature increases conduction-electron number, lowering resistance and increasing current.

In a potential divider, translate this resistance change only after identifying the output position. As an NTC thermistor warms, p.d. across it falls if it is the output resistor, while p.d. across the fixed series resistor rises.

When illumination increases, more light energy arrives at an LDR and more electrons become available as conduction carriers. The increased carrier number allows a larger current for the same p.d., so the LDR's resistance decreases.

In lower illumination, fewer conduction electrons are available and resistance is higher. This carrier-number model connects the environmental input to the electrical response rather than treating the LDR as a switch with only two states.

In a fixed-voltage series circuit, lower LDR resistance lowers total resistance and raises current. In a potential divider, output across the LDR falls with illumination, while output across the fixed resistor rises.