Physics important questions

(a) The principle of a potentiometer is that when a steady current flows through a uniform wire, the potential drop across any portion is proportional to its length, $V \propto \ell$. To compare two emfs, each cell is in turn balanced ag...

(a) Principle of the potentiometer. When a steady current flows through a wire of uniform cross section and uniform material, the potential difference across any portion of the wire is directly proportional to the length of that portion:

$$V\propto l,\qquad V=kl$$

where $k$ is the potential gradient, the fall of potential per unit length. This follows because $V=IR$ and $R=\rho l/A$, so with $I$, $\rho$ and $A$ all constant, $V=\left(\dfrac{I\rho}{A}\right)l$.

Internal resistance of a cell. The potentiometer wire $AB$ is fed by a driver cell through a key, and the cell under test, of emf $E$ and internal resistance $r$, is connected with its positive terminal to $A$ and through a galvanometer to the jockey. A resistance box $R$ with a key is joined across the test cell:

   +|i----[ key ]----------------------------
    |                                        |
   === driver                                |
    |                                        |
    A o======================================o B   (uniform wire)
      |                  |
      |              jockey ---- (G) ---- +|i---- test cell (E, r)
      |                                       |
      +---------------------------------------+
                         |
                    [ R ] [ key K ]  across the test cell

With the key $K$ open no current is drawn from the test cell, so the balancing length $l_1$ measures its full emf:

$$E=k,l_1$$

With $K$ closed, the cell sends current through $R$, and the balance point now measures only the terminal potential difference $V$:

$$V=k,l_2$$

For the closed circuit $E=I(R+r)$ and $V=IR$, so

$$ \begin{aligned} \frac{E}{V} &= \frac{R+r}{R} \ &= \frac{l_1}{l_2} \end{aligned} $$

which rearranges to

$$r=R\left(\frac{l_1-l_2}{l_2}\right)$$

Measuring $l_1$, $l_2$ and $R$ therefore gives the internal resistance. The method is a null method, so at balance no current is drawn from the cell being measured and the result is free from the error that a voltmeter would introduce.

(b) Thermoelectric effect. When two different metals are joined to form a closed circuit and the two junctions are kept at different temperatures, an emf is set up and a current flows round the circuit. This is the thermoelectric or Seebeck effect, and the pair of metals is a thermocouple. The effect is reversible: interchanging the hot and cold junctions reverses the current.

Variation with the temperature of the hot junction. With the cold junction held at $0^\circ\text{C}$ and the hot junction at $\theta$, the thermo emf is not linear but parabolic:

$$E=\alpha\theta+\tfrac{1}{2}\beta\theta^{2}$$

where $\alpha$ and $\beta$ are constants for the pair. As $\theta$ rises from zero the emf increases, but at a falling rate, until it reaches a maximum at the neutral temperature $\theta_n$, which for a given couple is fixed (about $270^\circ\text{C}$ for a copper-iron couple). Beyond $\theta_n$ the emf decreases, falls back to zero at the temperature of inversion $\theta_i$, and on further heating reverses its direction. Since the curve is a parabola about $\theta_n$, the neutral temperature lies midway between the cold junction temperature and the inversion temperature:

$$\theta_n=\frac{\theta_c+\theta_i}{2}$$

so with the cold junction at $0^\circ\text{C}$, $\theta_i=2\theta_n$. Differentiating gives the thermoelectric power $\dfrac{dE}{d\theta}=\alpha+\beta\theta$, which is zero at the neutral temperature, as the maximum requires.

(c) Magnetic elements of the earth. These are the three quantities needed to specify the earth's magnetic field completely at a place:

1. Declination ($\theta$), the angle between the geographic meridian and the magnetic meridian at that place.
2. Dip or inclination ($\delta$), the angle which the earth's total field makes with the horizontal.
3. Horizontal component ($H$) of the earth's total field $B$, with $H=B\cos\delta$ and the vertical component $V=B\sin\delta$, so that $\tan\delta=V/H$.

Proof of the relation. The true dip $\delta$ is observed only when the dip circle is set in the magnetic meridian. If instead the plane of the dip circle is turned through an angle $\alpha$ from the magnetic meridian, the vertical component $V$ is unaffected, but only the resolved part of the horizontal component acts in that plane, namely $H\cos\alpha$. The apparent dip $\delta_1$ in that plane satisfies

$$\tan\delta_1=\frac{V}{H\cos\alpha}$$

Let the second dip circle be set in the plane perpendicular to the first, so its plane makes an angle $(90^\circ-\alpha)$ with the magnetic meridian. Then

$$ \begin{aligned} \tan\delta_2 &= \frac{V}{H\cos(90^\circ-\alpha)} \ &= \frac{V}{H\sin\alpha} \end{aligned} $$

Taking reciprocals and squaring both results,

$$ \begin{aligned} \cot^2\delta_1 &= \frac{H^2\cos^2\alpha}{V^2} \ \qquad \cot^2\delta_2 &= \frac{H^2\sin^2\alpha}{V^2} \end{aligned} $$

Adding them,

$$ \begin{aligned} \cot^2\delta_1+\cot^2\delta_2 &= \frac{H^2}{V^2}\left(\cos^2\alpha+\sin^2\alpha\right) \ &= \frac{H^2}{V^2} \end{aligned} $$

But for the true dip $\tan\delta=V/H$, so $\cot^2\delta=H^2/V^2$. Hence

$$\cot^2\delta=\cot^2\delta_1+\cot^2\delta_2$$

which is the required result. It is useful in practice because it gives the true dip from two measurements in mutually perpendicular planes, without having to locate the magnetic meridian first.

(d) Series LCR circuit. Let an alternating emf $e=e_0\sin\omega t$ drive a current $I=I_0\sin(\omega t-\phi)$ through $R$, $L$ and $C$ in series. The same current passes through all three, so the current is taken as the reference in the phasor diagram.

The potential difference across the resistance is in phase with the current, that across the inductor leads it by $90^\circ$, and that across the capacitor lags it by $90^\circ$:

$$ \begin{aligned} V_R &= IR \ \qquad V_L &= IX_L \ &= I\omega L \ \qquad V_C &= IX_C \ &= \frac{I}{\omega C} \end{aligned} $$

Since $V_L$ and $V_C$ are opposite in phase, they combine to a single phasor of magnitude $|V_L-V_C|$ at right angles to $V_R$. The applied voltage is the resultant of these two perpendicular phasors:

$$ \begin{aligned} V &= \sqrt{V_R^{2}+(V_L-V_C)^{2}} \ &= I\sqrt{R^{2}+(X_L-X_C)^{2}} \end{aligned} $$

The impedance is the ratio of the applied voltage to the current, so

$$ \begin{aligned} Z &= \frac{V}{I} \ &= \sqrt{R^{2}+(X_L-X_C)^{2}} \ &= \sqrt{R^{2}+\left(\omega L-\frac{1}{\omega C}\right)^{2}} \end{aligned} $$

The phase angle $\phi$ between the applied voltage and the current is the angle of that resultant with $V_R$:

$$ \begin{aligned} \tan\phi &= \frac{V_L-V_C}{V_R} \ &= \frac{X_L-X_C}{R} \ \phi &= \tan^{-1}\left(\frac{\omega L-\dfrac{1}{\omega C}}{R}\right) \end{aligned} $$

If $X_L>X_C$ the circuit is inductive and the current lags the voltage; if $X_C>X_L$ it is capacitive and the current leads; and if $X_L=X_C$ then $\phi=0$ and $Z=R$, which is the condition of resonance, when the current is a maximum.

(a) In a metal the free electrons move about randomly with high thermal speeds, but their average velocity is zero, so there is no net current. When a field $E$ is applied, each electron acquires a small drift velocity $v_d$ directed opposite to $E$ in the intervals between collisions. In a time $dt$ the charge crossing an area $A$ is $dq = neAv_d,dt$, so the current is $I = neAv_d$, and dividing by the area gives the current density,

$$ \begin{aligned} J &= \frac{I}{A} \ &= nev_d, \end{aligned} $$

where $n$ is the number density of free electrons and $e$ is the electronic charge.

(b) The Seebeck effect is the setting up of an emf, and hence a current, in a closed circuit of two dissimilar metals when its two junctions are kept at different temperatures. Keeping the cold junction at $0^\circ$C and gradually raising the hot-junction temperature $\theta$, the thermo-emf first increases, reaches a maximum at the neutral temperature $\theta_n$, then decreases and becomes zero at the temperature of inversion $\theta_i$, which lies symmetrically about $\theta_n$, and reverses in sign beyond that. The graph of $E$ against $\theta$ is therefore a parabola, $E = \alpha\theta + \tfrac{1}{2}\beta\theta^2$.

(c) The Biot-Savart law states that the magnetic field due to a current element $I,d\vec l$ at a point a distance $r$ away is $dB = \dfrac{\mu_0}{4\pi}\dfrac{I,dl\sin\theta}{r^2}$. For a circular coil of radius $a$ and $N$ turns, at a point on the axis a distance $x$ from the centre, the components of $dB$ perpendicular to the axis cancel while the axial components of all the elements add up, giving

$$B = \frac{\mu_0 N I a^2}{2,(a^2 + x^2)^{3/2}}.$$

At the centre of the coil, where $x = 0$, this reduces to $B = \dfrac{\mu_0 N I}{2a}$.

(d) Faraday's law states that the induced emf equals the negative rate of change of flux linkage, $\varepsilon = -N\dfrac{d\Phi}{dt}$. For a coil of $N$ turns and area $A$ rotating with angular velocity $\omega$ in a uniform field $B$, the flux through it is $\Phi = BA\cos\omega t$, so differentiating gives

$$ \begin{aligned} \varepsilon &= -N\frac{d\Phi}{dt} \ &= NBA,\omega\sin\omega t \ &= \varepsilon_0\sin\omega t, \ \qquad \varepsilon_0 &= NBA\omega. \end{aligned} $$

This is the sinusoidal emf of an a.c. generator, with peak value $\varepsilon_0 = NBA\omega$.

(a) A Wheatstone bridge is an arrangement of four resistances $P, Q, R, S$ forming a quadrilateral, with a galvanometer connected across one diagonal and a cell across the other.

At balance no current flows through the galvanometer. Applying Kirchhoff's voltage law to the two loops with $I_g = 0$ gives $I_1 P = I_2 R$ for the upper loop and $I_1 Q = I_2 S$ for the lower loop. Dividing one relation by the other, the currents cancel and we are left with the balance condition,

$$\frac{P}{Q} = \frac{R}{S}$$

(b) The Seebeck effect is the appearance of a thermo-emf, and hence a current, in a closed circuit of two dissimilar metals when its two junctions are held at different temperatures.

Keeping the cold junction fixed and raising the hot-junction temperature, the thermo-emf first increases, reaches a maximum at the neutral temperature $\theta_n$, then decreases, and finally becomes zero and reverses at the temperature of inversion $\theta_i$. This variation follows a parabolic law,

$$E = a\theta + \tfrac12 b\theta^2$$

(c) Conductor 1 carrying current $I_1$ sets up a magnetic field at the location of conductor 2 (carrying $I_2$ at a distance $d$),

$$B_1 = \frac{\mu_0 I_1}{2\pi d}$$

This field exerts a force on conductor 2, so the force per unit length is

$$ \begin{aligned} \frac{F}{l} &= B_1 I_2 \ &= \frac{\mu_0 I_1 I_2}{2\pi d} \end{aligned} $$

One ampere is defined as that steady current which, flowing in two infinitely long parallel conductors placed 1 m apart in vacuum, produces a force of $2\times10^{-7}\ \text{N/m}$ between them.

(d) Let the current through the series R-L combination be $I = I_0\sin\omega t$. The voltage across the resistor is in phase with the current, $V_R = I_0 R\sin\omega t$, while the voltage across the inductor leads it by $90^\circ$, $V_L = I_0 X_L\sin(\omega t + 90^\circ)$. Adding these two out-of-phase voltages, the applied voltage is

$$ \begin{aligned} V &= V_0\sin(\omega t + \phi) \ \qquad V_0 &= I_0\sqrt{R^2 + X_L^2} \ \quad X_L &= \omega L \end{aligned} $$

so the impedance is $Z = \sqrt{R^2 + \omega^2L^2}$. Because of the inductor, the voltage leads the current by the phase angle

$$\phi = \tan^{-1}!\frac{\omega L}{R}$$

(a) The resistance follows from the power rating, since $R = \dfrac{V^2}{P}$: $$ \begin{aligned} R &= \frac{12^2}{25} \ &= \frac{144}{25} \ &= 5.76\ \Omega. \end{aligned} $$ With $A = 10\ \text{mm}^2 = 1\times10^{-5}\ \text{m}^2$, $\rh...

(a) The resistance of a conductor varies with temperature as $R_t=R_0(1+\alpha t)$, where $R_0$ is the resistance at $0^\circ\text{C}$ and $\alpha$ the temperature coefficient. The two measurements give

$$ \begin{aligned} 3.15 &= R_0(1+20\alpha) \ 3.75 &= R_0(1+100\alpha) \end{aligned} $$

Dividing the second equation by the first eliminates $R_0$:

$$ \begin{aligned} \frac{3.75}{3.15} &= \frac{1+100\alpha}{1+20\alpha} \ 1.19048(1+20\alpha) &= 1+100\alpha \ 1.19048+23.8095\alpha &= 1+100\alpha \ 0.19048 &= 76.1905,\alpha \ \alpha &= 2.5\times10^{-3}\ ^\circ\text{C}^{-1} \end{aligned} $$

Substituting this back into the first equation,

$$ \begin{aligned} R_0 &= \frac{3.15}{1+20\times2.5\times10^{-3}} \ &= \frac{3.15}{1.05} \ &= 3.0\ \Omega \end{aligned} $$

So the temperature coefficient is $\mathbf{2.5\times10^{-3}\ ^\circ C^{-1}}$ and the resistance at $0^\circ\text{C}$ is $\mathbf{3\ \Omega}$.

(b) The straight wire produces a magnetic field at the position of the electron of magnitude

$$ \begin{aligned} B &= \frac{\mu_0 I}{2\pi a} \ &= \frac{4\pi\times10^{-7}\times5}{2\pi\times0.10} \ &= \frac{2\times10^{-7}\times5}{0.10} \ &= 1\times10^{-5}\ \text{T} \end{aligned} $$

This field circles the wire, so at the electron it points perpendicular to the wire. The electron travels parallel to the wire, so its velocity is at right angles to $\vec{B}$ and $\theta=90^\circ$. The magnetic force on it is

$$ \begin{aligned} F &= evB\sin\theta \ &= 1.6\times10^{-19}\times10^{6}\times1\times10^{-5}\times 1 \ F &= 1.6\times10^{-18}\ \text{N} \end{aligned} $$

The force experienced by the electron is $\mathbf{1.6\times10^{-18}\ N}$, directed perpendicular to both the wire and the velocity, that is towards or away from the wire depending on the sense of the current and the direction of motion.

(c) For a coil rotating in a uniform field the induced emf is $e=NBA\omega\sin\omega t$, whose maximum value is

$$e_0=NBA\omega$$

The area of the square coil is

$$ \begin{aligned} A &= (0.10)^2 \ &= 0.01\ \text{m}^2 \end{aligned} $$

and the angular speed corresponding to $500$ revolutions per minute is

$$ \begin{aligned} \omega &= 2\pi\times\frac{500}{60} \ &= 2\pi\times8.333 \ &= 52.36\ \text{rad/s} \end{aligned} $$

Therefore

$$ \begin{aligned} e_0 &= 100\times0.5\times0.01\times52.36 \ e_0 &= 26.18\ \text{V} \end{aligned} $$

The maximum emf produced in the coil is about $\mathbf{26.2\ V}$.

(a) The three branches all share the same node voltage $V$, which appears across the $7\ \Omega$ resistor. Since the current through it is $1.80\ \text{A}$,

$$ \begin{aligned} V &= 1.80\times7 \ &= 12.6\ \text{V} \end{aligned} $$

The current fed into the top node by the $24\ \text{V}$ branch is

$$ \begin{aligned} I_1 &= \frac{24 - V}{3} \ &= \frac{24 - 12.6}{3} \ &= 3.8\ \text{A} \end{aligned} $$

Applying Kirchhoff's current law at the node, $I_1 + I_2 = I_{7\Omega}$, the current in the $E$ branch is

$$ \begin{aligned} I_2 &= 1.80 - 3.8 \ &= -2.0\ \text{A} \end{aligned} $$

the negative sign meaning it actually flows into the $E$ branch. For that branch $V = E - I_2(2)$, so

$$ \begin{aligned} E &= V + 2I_2 \ &= 12.6 + 2(-2.0) \ &= 8.6\ \text{V} \end{aligned} $$

Therefore the required emf is $E = 8.6\ \text{V}$.

(b) For the rod to be self-supporting, the upward magnetic force must balance its weight, $BIL = mg$. With $L = 0.20\ \text{m}$, $m = 0.030\ \text{kg}$, $I = 2\ \text{A}$ and $g = 9.8\ \text{m s}^{-2}$,

$$ \begin{aligned} B &= \frac{mg}{IL} \ &= \frac{(0.030)(9.8)}{(2)(0.20)} \ &= \frac{0.294}{0.40} \ &= 0.735\ \text{T} \end{aligned} $$

(c) Since the voltage across the inductor equals that across the resistor, $V_L = V_R$, which means $X_L = R = 40\ \Omega$. The supply voltage is $V = \sqrt{V_R^2 + V_L^2} = V_R\sqrt{2}$, so

$$ \begin{aligned} V_L &= V_R \ &= \frac{V}{\sqrt{2}} \ &= \frac{50}{\sqrt{2}} \ &= 35.4\ \text{V} \end{aligned} $$

Using $X_L = 2\pi f L$, the supply frequency is

$$ \begin{aligned} f &= \frac{X_L}{2\pi L} \ &= \frac{40}{2\pi(0.1)} \ &= 63.7\ \text{Hz} \end{aligned} $$

Therefore the voltage across the inductor is $35.4\ \text{V}$ and the frequency is $63.7\ \text{Hz}$.

(a) A $6\ \text{V}$ battery of internal resistance $r_1 = 0.5\ \Omega$ is in parallel with a $10\ \text{V}$ battery of internal resistance $r_2 = 1\ \Omega$, and the pair feeds a $12\ \Omega$ resistor. Let $V$ be the p.d. across the parallel combination, which is also the p.d. across the $12\ \Omega$ resistor. Then the battery currents are $I_1 = \dfrac{6-V}{0.5}$ and $I_2 = \dfrac{10-V}{1}$, and by Kirchhoff's current law $I_1 + I_2 = \dfrac{V}{12}$.

$$\frac{6-V}{0.5} + \frac{10-V}{1} = \frac{V}{12}$$

Clearing the terms,

$$ \begin{aligned} (12 - 2V) + (10 - V) &= \frac{V}{12} \ 22 - 3V &= \frac{V}{12} \quad\Rightarrow\quad 264 \ &= 37V \quad\Rightarrow\quad V \ &= 7.14\ \text{V} \end{aligned} $$

Substituting back gives the branch currents,

$$ \begin{aligned} I_1 &= \frac{6 - 7.14}{0.5} \ &= -2.27\ \text{A}\quad (\text{this battery is being charged}) \ I_2 &= \frac{10 - 7.14}{1} \ &= 2.87\ \text{A}, \ \qquad \text{external current} &= 0.60\ \text{A} \end{aligned} $$

So the $6\ \text{V}$ battery carries $2.27\ \text{A}$ in reverse (it is being charged) while the $10\ \text{V}$ battery delivers $2.87\ \text{A}$.

(b) A galvanometer of $50$ turns and $10\ \Omega$ is replaced by one of $100$ turns and $50\ \Omega$. Current sensitivity is proportional to the number of turns, $S_I \propto N$, so it changes by

$$\frac{100}{50} = 2 \quad (\text{doubles})$$

Voltage sensitivity is proportional to $N/R$, so it changes by

$$ \begin{aligned} \frac{100/50}{50/10} &= \frac{2}{5} \ &= 0.4\quad (\text{becomes 0.4 times}) \end{aligned} $$

Thus the current sensitivity doubles while the voltage sensitivity drops to $0.4$ times its former value.

(c) The solenoid has $N = 1000$ turns, cross-section $A = 2\times10^{-3}\ \text{m}^2$, carries $2\ \text{A}$, and produces a flux density $B = 52\times10^{-3}\ \text{T}$. Its self-inductance follows from $L = \dfrac{N\Phi}{I} = \dfrac{NBA}{I}$.

$$ \begin{aligned} L &= \frac{1000\times52\times10^{-3}\times2\times10^{-3}}{2} \ &= \frac{0.104}{2} \ &= 0.052\ \text{H} \end{aligned} $$

So the self-inductance is $0.052\ \text{H}$, or $52\ \text{mH}$.

(a) As a voltmeter the coil has resistance $R_{coil} = 50\ \Omega$ with a series resistor $R_{series} = 1.15\ \text{k}\Omega = 1150\ \Omega$ and reads full scale at $12\ \text{V}$. The full-scale current through the movement follows from Ohm's law across the whole voltmeter branch.

$$ \begin{aligned} I_g &= \frac{V}{R_{coil}+R_{series}} \ &= \frac{12}{50+1150} \ &= 0.01\ \text{A} \end{aligned} $$

To use it as an ammeter reading up to $I = 2\ \text{A}$, a shunt $S$ is placed in parallel with the coil and carries the excess current $(I - I_g)$. Since the coil and shunt share the same voltage, $I_g R_{coil} = (I - I_g)S$, which gives

$$ \begin{aligned} S &= \frac{I_g R_{coil}}{I - I_g} \ &= \frac{0.01\times50}{2 - 0.01} \ &= 0.251\ \Omega \end{aligned} $$

So the required shunt resistance is about $0.251\ \Omega$.

(b) For the copper slab the thickness along the field is $t = 2\ \text{mm} = 2\times10^{-3}\ \text{m}$, the width is $1.50\ \text{cm}$, the field is $B = 0.40\ \text{T}$, the current is $I = 75\ \text{A}$, and the measured Hall voltage is $V_H = 0.81\ \mu\text{V}$, with electron charge $e = 1.6\times10^{-19}\ \text{C}$. The Hall voltage is related to the carrier concentration by $V_H = \dfrac{BI}{n e t}$, so rearranging for $n$,

$$n = \frac{BI}{V_H e t}$$

Substituting the values,

$$ \begin{aligned} n &= \frac{0.40\times75}{(0.81\times10^{-6})(1.6\times10^{-19})(2\times10^{-3})} \ &= \frac{30}{2.592\times10^{-28}} \ &= 1.16\times10^{29}\ \text{m}^{-3} \end{aligned} $$

So the mobile-electron concentration is about $1.16\times10^{29}\ \text{m}^{-3}$.

(c) The coil has $N = 100$ turns, sides $15\times10\ \text{cm}$, spins at $300\ \text{rpm}$ in a field $B = 0.6\ \text{T}$. First find the angular frequency and the area.

$$ \begin{aligned} \omega &= 2\pi\frac{300}{60} \ &= 31.42\ \text{rad/s}, \ \qquad A &= 0.15\times0.10 \ &= 0.015\ \text{m}^2 \end{aligned} $$

The peak emf of a rotating coil is $E_0 = NBA\omega$, so

$$ \begin{aligned} E_0 &= 100\times0.6\times0.015\times31.42 \ &= 28.3\ \text{V} \end{aligned} $$

The maximum emf is therefore about $28.3\ \text{V}$.

Ohm's law states that, at constant temperature (and under unchanged physical conditions), the current $I$ flowing through a conductor is directly proportional to the potential difference $V$ across its ends, $$V \propto I \quad\Rightarro...

a) We first balance the emf of the cell with the circuit open (the key to $R$ open), where the balancing length is $l_1$, so $E = kl_1$ with $k$ the potential gradient of the wire. A known resistance $R$ is then connected across the cell, and now the potentiometer balances the terminal p.d. $V$ at a length $l_2$, so $V = kl_2$. For the cell we have $E = I(R + r)$ and $V = IR$, and taking the ratio,

$$ \begin{aligned} \frac{E}{V} &= \frac{R + r}{R} \ &= \frac{l_1}{l_2}, \end{aligned} $$

which rearranges to

$$\boxed{r = R\left(\frac{l_1 - l_2}{l_2}\right)}.$$

b) A galvanometer is converted into an ammeter by connecting a low resistance, called a shunt $S$, in parallel with it. If $I_g$ is the full-scale current of the galvanometer whose coil resistance is $G$, and $I$ is the range required, then the shunt must carry the remaining current $(I - I_g)$. Since the coil and shunt are in parallel they have the same p.d. across them,

$$I_g G = (I - I_g)S \quad\Rightarrow\quad \boxed{S = \frac{I_g G}{I - I_g}}.$$

The shunt diverts most of the current, so the instrument can read up to $I$ while only $I_g$ actually passes through the coil.

A potentiometer works on a very simple idea. If a steady current is maintained through a wire of uniform cross-section and uniform composition, the fall of potential along the wire is uniform, so the potential difference across any portion of the wire is directly proportional to the length of that portion. Writing this for a portion of length $l$,

$$V \propto l \quad \text{that is} \quad V = kl$$

where $k$ is the potential gradient of the wire, the fall of potential per unit length. Because the balance point is located by a null deflection of the galvanometer, no current is drawn from the source under test at balance, so a potentiometer measures an emf or a potential difference without loading the source. This is its advantage over an ordinary voltmeter.

To find the internal resistance of a cell, the cell of emf $E$ and internal resistance $r$ is connected to the potentiometer wire through a galvanometer and a jockey, and a resistance box $R$ is joined across the cell through a key so that it can serve as a shunt. With the key open the cell sends no current through itself, so the wire balances the full emf of the cell. If $l_1$ is the balancing length in this condition,

$$E = k l_1$$

The key is now closed so that the cell drives a current through the shunt $R$, and the balance point is found afresh at a length $l_2$. What the wire balances now is the terminal potential difference $V$ of the cell, so

$$V = k l_2$$

The current sent through the circuit is $I = E/(R + r)$ and the terminal potential difference across the shunt is $V = IR$, so on dividing,

$$ \begin{aligned} \frac{E}{V} &= \frac{I(R + r)}{IR} \ &= \frac{R + r}{R} \end{aligned} $$

Replacing the emf and the potential difference by the two balancing lengths,

$$\frac{l_1}{l_2} = \frac{R + r}{R}$$

and rearranging this gives the working expression

$$r = R\left(\frac{l_1 - l_2}{l_2}\right)$$

So with a known shunt resistance $R$ and the two balancing lengths, the internal resistance of the cell follows at once. Note that $l_1$ is always greater than $l_2$, because the terminal potential difference of a cell that is delivering current is less than its emf by the drop $Ir$ inside the cell.

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The wire is a uniform cylinder, so the whole problem follows from the resistance formula for a uniform conductor once the area of cross section has been worked out. We are given the length $L = 1$ m, the diameter $d = 1$ mm $= 1 \times 1...

The cell has an emf of 1.5 V, and when a 25 ohm resistor is joined across it the terminal potential difference is only 1.25 V. The emf is the energy the cell supplies to each coulomb of charge, while the terminal p.d. is the part of that...

The lamp is rated 60 W at 220 V, which means it is meant to have 220 V across it and, at that voltage, to draw the current that goes with the rating, $$ \begin{aligned} I &= \frac{P}{V} \ &= \frac{60}{220} \ &= 0.2727\ \text{A} \end{al...

Kirchhoff gave two rules for networks that cannot be reduced by series and parallel combinations. His junction rule states that the algebraic sum of the currents meeting at any junction is zero, currents flowing towards the junction bein...

In a wire of uniform cross section carrying a steady current the current density is the same everywhere across the section, and the electric field that keeps the free electrons drifting is tied to that current density by the resistivity ...

A potentiometer measures a potential difference by balancing it against the fall of potential along a uniform resistance wire that carries a steady current from a driver cell. Here the potential gradient along the wire is handed to us as...

The heater element and the connecting wires are joined in series, so exactly the same current $I$ flows through both of them, and the heat developed in a time $t$ is given by Joule's law,

$$H = I^2Rt$$

Since $I$ and $t$ are common, the heating depends only on the resistance $R$. The element is a long thin coil of nichrome, a material of high resistivity, so its resistance is large and a great deal of heat appears in it. The leads are short, thick copper wires of very small resistance, so hardly any heat is produced in them, and the little they do produce passes easily to the surroundings. The element therefore becomes red hot while the wires stay almost cool.

The value of the resistance obtained from a given set of identical wires depends entirely on how they are connected, so the question is really about which combination gives the extreme values.

The greatest resistance is obtained by joining all $n$ wires in series, because in a series combination the same current passes through every wire and the individual resistances simply add:

$$ \begin{aligned} & R_{max} = R + R + \cdots + R \ \ & (n\ \text{terms}) = nR \end{aligned} $$

The least resistance is obtained by joining all $n$ wires in parallel, because the current then divides among $n$ identical paths and the reciprocals of the resistances add, giving $\dfrac{1}{R_{min}} = \dfrac{n}{R}$, that is

$$ R_{min} = \frac{R}{n} $$

Dividing the first result by the second, the individual resistance $R$ cancels:

$$ \begin{aligned} \frac{R_{max}}{R_{min}} &= \frac{nR}{R/n} \ &= n^{2} \end{aligned} $$

Therefore the ratio of the maximum to the minimum resistance obtainable from $n$ equal wires is $n^{2} : 1$. For instance, four wires of 6 $\Omega$ each give 24 $\Omega$ in series and 1.5 $\Omega$ in parallel, a ratio of 16, which is indeed $4^{2}$.