Electrical Circuits

a) Kirchhoff's laws

Kirchhoff's first law (junction law / current law): The algebraic sum of the currents meeting at a junction is zero, $\sum I = 0$. In other words, the total current entering a junction equals the total current leaving it. It follows from conservation of charge.

Kirchhoff's second law (loop law / voltage law): In any closed loop of a circuit, the algebraic sum of the products of current and resistance ($IR$) equals the algebraic sum of the emfs in that loop, $\sum IR = \sum E$. It follows from conservation of energy.

b) Balanced condition of a Wheatstone bridge

              I1        B        I1 - Ig
        +-----[ P ]-----+-----[ Q ]-----+
        |               |               |
   I -->|              (G) Ig           |--> I
      A |               |               | C
        +-----[ R ]-----+-----[ S ]-----+
        |    I - I1     D  I - I1 + Ig  |
        |                               |
        +-----------| |------/ ---------+
                     E       K

Setup: Four resistances $P$, $Q$, $R$, $S$ form the arms AB, BC, AD, DC. A galvanometer of resistance $G$ is connected between B and D, and a cell $E$ with key K between A and C. Let a current $I$ enter at A, $I_1$ flow through $P$, and $I_g$ flow through the galvanometer from B to D.

1. Junction law gives the currents in the other arms:

   • in $Q$ (B to C): $I_1 - I_g$
   • in $R$ (A to D): $I - I_1$
   • in $S$ (D to C): $I - I_1 + I_g$

2. Loop law for loop ABDA:

$$I_1 P + I_g G - (I - I_1)R = 0 \qquad (1)$$

3. Loop law for loop BCDB:

$$(I_1 - I_g)Q - (I - I_1 + I_g)S - I_g G = 0 \qquad (2)$$

4. Balance condition: $R$ (or another arm) is adjusted until the galvanometer shows no deflection, so $I_g = 0$. Then (1) and (2) become

$$I_1 P = (I - I_1)R \qquad (3)$$ $$I_1 Q = (I - I_1)S \qquad (4)$$

5. Dividing (3) by (4):

$$\boxed{\dfrac{P}{Q} = \dfrac{R}{S}}$$

This is the balanced condition of a Wheatstone bridge. If three resistances are known, the fourth can be found: $S = \dfrac{Q}{P}R$.

(a) $10^{-2}$ V/m

The potential gradient is $k = \dfrac{V}{l} = \dfrac{IR}{l}$. Since $R = \dfrac{\rho l}{A}$,

$$k = \frac{I\rho}{A} = \frac{0.1 \times 10^{-7}}{10^{-6}} = 10^{-2}\ \text{V/m}$$

(a) Significance of the first law

Kirchhoff's first law ($\sum I = 0$ at a junction) shows that charge is conserved: charge cannot pile up at, or be created at, a junction, so the current flowing in equals the current flowing out. It lets us write down the current in every branch of a network.

(b) Kirchhoff's second law

Statement: In any closed loop of an electrical circuit, the algebraic sum of the products of current and resistance equals the algebraic sum of the emfs in the loop:

$$\sum IR = \sum E$$

Explanation: The law follows from conservation of energy. A unit charge taken once round a closed loop comes back to the same point, so its net change in potential energy is zero. The energy it gains from the emfs equals the energy it loses in the resistances.

         E1 (+ -)          R1
    +-------| |--------/\/\/\/----+
    |                             |
    |        I  (clockwise) -->   |
    |                             |
    +----/\/\/\/--------| |-------+
            R2       (- +) E2

In the loop above, $E_2$ is connected so that it opposes $E_1$. Going round clockwise with current $I$:

$$IR_1 + IR_2 = E_1 - E_2$$

Sign rule: an $IR$ term is positive when we move along the current, and an emf is positive when we go through the cell from $-$ to $+$.

(c) Using the laws to find an unknown resistance (Wheatstone bridge)

              I1        B        I1 - Ig
        +-----[ P ]-----+-----[ Q ]-----+
        |               |               |
   I -->|              (G) Ig           |--> I
      A |               |               | C
        +-----[ R ]-----+-----[ S ]-----+
        |    I - I1     D  I - I1 + Ig  |
        |                               |
        +-----------| |------/ ---------+
                     E       K

Setup: Four resistances $P$, $Q$, $R$, $S$ form the arms AB, BC, AD, DC. A galvanometer of resistance $G$ is connected between B and D, and a cell $E$ with key K between A and C. Let a current $I$ enter at A, $I_1$ flow through $P$, and $I_g$ flow through the galvanometer from B to D.

1. Junction law gives the currents in the other arms:

   • in $Q$ (B to C): $I_1 - I_g$
   • in $R$ (A to D): $I - I_1$
   • in $S$ (D to C): $I - I_1 + I_g$

2. Loop law for loop ABDA:

$$I_1 P + I_g G - (I - I_1)R = 0 \qquad (1)$$

3. Loop law for loop BCDB:

$$(I_1 - I_g)Q - (I - I_1 + I_g)S - I_g G = 0 \qquad (2)$$

4. Balance condition: $R$ (or another arm) is adjusted until the galvanometer shows no deflection, so $I_g = 0$. Then (1) and (2) become

$$I_1 P = (I - I_1)R \qquad (3)$$ $$I_1 Q = (I - I_1)S \qquad (4)$$

5. Dividing (3) by (4):

$$\boxed{\dfrac{P}{Q} = \dfrac{R}{S}}$$

This is the balanced condition of a Wheatstone bridge. If three resistances are known, the fourth can be found: $S = \dfrac{Q}{P}R$.

In a meter bridge with a known resistance $R$ in the left gap, an unknown $X$ in the right gap, and the balance point at length $l$ cm from the left end:

$$\frac{R}{X} = \frac{l}{100 - l} \quad\Rightarrow\quad \mathbf{X = R\,\dfrac{100 - l}{l}}$$

(d) Meter bridge

A meter bridge (slide-wire bridge) is a practical form of the Wheatstone bridge. It uses a 1 m long uniform wire stretched along a metre scale. The two parts of the wire on either side of the jockey form two arms of the bridge, and the known and unknown resistances form the other two. It is used to find an unknown resistance.

Material: The bridge wire is made of constantan or manganin, which have high resistivity and a very low temperature coefficient of resistance. The connecting strips are thick copper strips, which have negligible resistance.

(a) Principle

When a steady current flows through a wire of uniform cross-section and uniform material, the potential difference across any length of the wire is directly proportional to that length:

$$V \propto l \quad\Rightarrow\quad V = kl$$

where $k$ is the potential gradient (p.d. per unit length).

(b) Why a potentiometer is preferred to a voltmeter

At the balance point no current is drawn from the cell, so the potentiometer measures the true emf. A voltmeter always draws some current, so it reads the terminal p.d. $V = E - Ir$, which is less than the emf. A potentiometer therefore acts like a voltmeter of infinite resistance.

(c) Numerical

Given:

• balancing length for the battery, $l_1 = 75$ cm
• emf of the standard cell, $E_2 = 1.02$ V
• balancing length for the standard cell, $l_2 = 50$ cm

Since $E = kl$ for the same potential gradient:

$$\frac{E_1}{E_2} = \frac{l_1}{l_2}$$ $$E_1 = E_2 \times \frac{l_1}{l_2} = 1.02 \times \frac{75}{50} = 1.53\ \text{V}$$

The emf of the battery is 1.53 V.

a) Balanced condition of a Wheatstone bridge

              I1        B        I1 - Ig
        +-----[ P ]-----+-----[ Q ]-----+
        |               |               |
   I -->|              (G) Ig           |--> I
      A |               |               | C
        +-----[ R ]-----+-----[ S ]-----+
        |    I - I1     D  I - I1 + Ig  |
        |                               |
        +-----------| |------/ ---------+
                     E       K

Setup: Four resistances $P$, $Q$, $R$, $S$ form the arms AB, BC, AD, DC. A galvanometer of resistance $G$ is connected between B and D, and a cell $E$ with key K between A and C. Let a current $I$ enter at A, $I_1$ flow through $P$, and $I_g$ flow through the galvanometer from B to D.

1. Junction law gives the currents in the other arms:

   • in $Q$ (B to C): $I_1 - I_g$
   • in $R$ (A to D): $I - I_1$
   • in $S$ (D to C): $I - I_1 + I_g$

2. Loop law for loop ABDA:

$$I_1 P + I_g G - (I - I_1)R = 0 \qquad (1)$$

3. Loop law for loop BCDB:

$$(I_1 - I_g)Q - (I - I_1 + I_g)S - I_g G = 0 \qquad (2)$$

4. Balance condition: $R$ (or another arm) is adjusted until the galvanometer shows no deflection, so $I_g = 0$. Then (1) and (2) become

$$I_1 P = (I - I_1)R \qquad (3)$$ $$I_1 Q = (I - I_1)S \qquad (4)$$

5. Dividing (3) by (4):

$$\boxed{\dfrac{P}{Q} = \dfrac{R}{S}}$$

This is the balanced condition of a Wheatstone bridge. If three resistances are known, the fourth can be found: $S = \dfrac{Q}{P}R$.

b) Shift of the balance point in the given potentiometer circuit

In the figure, the driving cell $E_0$ and the resistance box $R$ are in series with the wire AB. The cell $E$, the resistance $S$ and the galvanometer are in the secondary circuit. At balance,

$$E = kl, \qquad k = \frac{\text{p.d. across AB}}{L} = \frac{E_0}{R + R_{AB}}\cdot\frac{R_{AB}}{L}$$

so

$$l = \frac{E}{k}$$

i) $R$ increased: The current in the wire AB decreases, so the p.d. across AB and the potential gradient $k$ both decrease. A longer length is then needed to balance the same emf $E$ ($l = E/k$ increases). The balance point shifts towards B.

ii) $S$ decreased, $R$ constant: $S$ is in the galvanometer (secondary) circuit, and at balance no current flows through it, so there is no p.d. across it. Since $k$ does not change, the balance point does not shift (it stays at the same position). Changing $S$ only changes how sensitive the galvanometer is near balance.

a) Internal resistance of a cell using a potentiometer

        +------| | | |------+          E0 (driver)
        |                  |
        A====P'=======P====B          potentiometer wire
        |    ^        ^
        |    |  jockey (to G)
        a----E---r----b---(G)--> jockey
        |             |
        +---[ R ]--K--+               R in parallel with the cell

Let $k$ be the potential gradient of the wire. It stays constant because the driver current is steady.

1. Key K open: No current is drawn from the cell, so the p.d. balanced is the emf $E$. Balancing length $= l_1$ (point P):

$$E = k l_1 \qquad (1)$$

2. Key K closed: A current $I$ flows through $R$. The p.d. balanced is now the terminal p.d. $V$. Balancing length $= l_2$ (point P'):

$$V = k l_2 \qquad (2)$$

3. For the cell in the closed circuit,

$$E = I(R + r), \qquad V = IR$$ $$\Rightarrow \frac{E}{V} = \frac{R + r}{R} \quad\Rightarrow\quad r = R\left(\frac{E}{V} - 1\right)$$

4. Using (1) and (2), $\dfrac{E}{V} = \dfrac{l_1}{l_2}$, so

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

Since $l_1 > l_2$, $r$ comes out positive. Knowing $R$, $l_1$ and $l_2$ gives the internal resistance.

b) Converting a galvanometer into an ammeter

A galvanometer (resistance $G$, full-scale current $I_g$) is converted into an ammeter of range $0$ to $I$ by connecting a low resistance $S$ (shunt) in parallel with it.

          I        Ig
   o----->---+---->(G)----+------o
             |            |
             +---[ S ]----+
               I - Ig

The p.d. across $G$ and $S$ is the same:

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

The shunt carries most of the current, and the combined resistance $\dfrac{GS}{G+S}$ is very small, as an ammeter needs.

Given:

• emf of the driver cell, $E_0 = 3$ V (internal resistance taken as negligible)
• resistance from the box, $R = 290\ \Omega$
• resistance of the wire, $R_w = 10\ \Omega$
• length of the wire, $L = 4$ m
• balancing length, $l = 240$ cm $= 2.40$ m

Current in the driver circuit:

$$I = \frac{E_0}{R + R_w} = \frac{3}{290 + 10} = 0.01\ \text{A}$$

P.d. across the wire and potential gradient:

$$V_w = IR_w = 0.01 \times 10 = 0.1\ \text{V}$$ $$k = \frac{V_w}{L} = \frac{0.1}{4} = 0.025\ \text{V/m}$$

Unknown emf:

$$E = kl = 0.025 \times 2.40 = 0.060\ \text{V}$$

The unknown emf is $E = 0.06$ V (60 mV).

a) Type of material

• Graph A: superconductor. As it is cooled, its resistivity falls, and at the critical temperature $T_c$ it drops suddenly (vertically) to zero. Below $T_c$ it has no resistance at all.
• Graph B: normal metallic conductor (e.g. copper). Its resistivity falls smoothly as the temperature decreases. Even near 0 K it keeps a small residual resistivity caused by impurities and defects; it never becomes zero.

b) Potential gradient

Definition: The potential gradient is the fall of potential per unit length along the potentiometer wire, $k = \dfrac{V}{l}$. Its SI unit is V m$^{-1}$.

Given:

• length of the wire, $L = 10$ m
• resistance of the wire, $R_w = 20\ \Omega$
• emf of the battery, $E = 3$ V (internal resistance neglected)
• series resistance, $R = 10\ \Omega$

$$I = \frac{E}{R_w + R} = \frac{3}{20 + 10} = 0.1\ \text{A}$$ $$V_w = IR_w = 0.1 \times 20 = 2\ \text{V}$$ $$k = \frac{V_w}{L} = \frac{2}{10} = 0.2\ \text{V m}^{-1}$$

The potential gradient is 0.2 V/m.

(c) Energy

Kirchhoff's voltage law says the algebraic sum of potential differences around a closed loop is zero. This means that the energy gained by a unit charge from the sources equals the energy it loses in the elements, so the law is a statement of conservation of energy.

(a) Let the bridge have arms $P$ (AB), $Q$ (BC), $R$ (AD) and $S$ (DC), with a galvanometer between B and D and a cell between A and C. Let the currents be $I_1$ in $P$, $I_2$ in $Q$, $I_3$ in $R$, $I_4$ in $S$, and $I_g$ in the galvanometer.

For balance, $I_g = 0$, so B and D are at the same potential. Then $I_2 = I_1$ and $I_4 = I_3$.

Applying Kirchhoff's voltage law to loop ABDA:

$I_1P - I_3R = 0 \;\Rightarrow\; I_1P = I_3R$

Applying it to loop BCDB:

$I_2Q - I_4S = 0 \;\Rightarrow\; I_1Q = I_3S$

Dividing the first by the second:

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

This is the balanced condition of the Wheatstone bridge.

(b) The current in the wire (neglecting the battery's internal resistance) is

$I = \frac{2}{10+5} = 0.133\ \text{A}$

The potential drop across the wire is $V = IR = 0.133\times10 = 1.33$ V. The potential gradient is

$k = \frac{V}{L} = \frac{1.33}{4} \approx 0.33\ \text{V/m}$

(a) Let the bridge have arms $P$ (AB), $Q$ (BC), $R$ (AD) and $S$ (DC), with a galvanometer between B and D and a cell between A and C. At balance, no current flows through the galvanometer, so B and D are at the same potential, and the currents in the arms are $I_1$ (through $P$ and $Q$) and $I_3$ (through $R$ and $S$).

Kirchhoff's voltage law on loop ABDA gives $I_1P = I_3R$, and on loop BCDB gives $I_1Q = I_3S$. Dividing:

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

(b) A very low resistance is comparable with, or even smaller than, the resistance of the connecting wires and the contact resistance at the junctions. These extra resistances get added to the arm and cannot be accounted for, so they cause large percentage errors. In addition, the small resistance gives a very small deflection and a poor sensitivity of balance. Special methods such as the Kelvin double bridge or the Carey Foster bridge are used instead.

(a) Principle and "infinite resistance"

Principle: When a constant current flows through a wire of uniform cross-section and material, the p.d. across any length of the wire is directly proportional to that length: $V = kl$.

Why it is called a voltmeter of infinite resistance: At the balance point the galvanometer shows no deflection, so no current is drawn from the cell (or circuit) whose p.d. is being measured. An ideal voltmeter of infinite resistance would also draw zero current. So a potentiometer behaves like a voltmeter of infinite resistance and measures the true emf.

(b)(i) Values doubled and interchanged

Initially, $R$ is in the left gap, $X$ is in the right gap and $l = 20$ cm:

$$\frac{R}{X} = \frac{l}{100 - l} = \frac{20}{80} = \frac{1}{4} \quad\Rightarrow\quad X = 4R$$

After doubling and interchanging, $2X$ is in the left gap and $2R$ is in the right gap. Let the new balancing length be $l'$:

$$\frac{2X}{2R} = \frac{l'}{100 - l'} \quad\Rightarrow\quad 4 = \frac{l'}{100 - l'}$$ $$400 - 4l' = l' \quad\Rightarrow\quad l' = 80\ \text{cm}$$

The new balance point is 80 cm from A (20 cm from B). Doubling both resistances does not change their ratio; only the interchange moves the balance point.

(b)(ii) Galvanometer and battery interchanged

The balance point is not affected. The balance condition $\dfrac{P}{Q} = \dfrac{R}{S}$ stays the same when the galvanometer and the cell are interchanged (this is the reciprocal property of the bridge). The galvanometer still shows zero deflection at the same point J.

(a) Kirchhoff's laws

Kirchhoff's first law (junction law / current law): The algebraic sum of the currents meeting at a junction is zero, $\sum I = 0$. In other words, the total current entering a junction equals the total current leaving it. It follows from conservation of charge.

Kirchhoff's second law (loop law / voltage law): In any closed loop of a circuit, the algebraic sum of the products of current and resistance ($IR$) equals the algebraic sum of the emfs in that loop, $\sum IR = \sum E$. It follows from conservation of energy.

(b) Numerical

With $R$ in the left gap and $S$ in the right gap, at balance $\dfrac{R}{S} = \dfrac{l}{100 - l}$.

Case 1: null point at $l_1 = 60.0$ cm.

$$\frac{R}{S} = \frac{60}{40} = 1.5 \quad\Rightarrow\quad R = 1.5\,S \qquad (1)$$

Case 2: $5\ \Omega$ in series with $S$, null point at $l_2 = 50$ cm.

$$\frac{R}{S + 5} = \frac{50}{50} = 1 \quad\Rightarrow\quad R = S + 5 \qquad (2)$$

From (1) and (2):

$$1.5\,S = S + 5 \quad\Rightarrow\quad 0.5\,S = 5 \quad\Rightarrow\quad S = 10\ \Omega$$ $$R = 1.5 \times 10 = 15\ \Omega$$

So $R = 15\ \Omega$ and $S = 10\ \Omega$.

(b) Diagram B

Rearranging, $I_1 - I_2 = I_3 - I_4$ gives $I_1 + I_4 = I_2 + I_3$. By Kirchhoff's junction law, this means $I_1$ and $I_4$ enter the junction while $I_2$ and $I_3$ leave it, which is exactly what Diagram B shows.

(a) Why no balance point, and the correction

The negative terminal of the test cell is joined to A, but A is connected to the positive terminal of the driving cell. So the emf of the test cell adds to the p.d. along the wire instead of opposing it. The meter deflects the same way wherever the jockey touches, so no null point is found.

Change needed: Reverse the test cell so that its positive terminal is connected to A (like terminals of the driving cell and the test cell joined at the same end). The emf of the driving cell must also be greater than that of the test cell. For best results, a sensitive centre-zero galvanometer should be used in place of the voltmeter.

(b) Recognizing the balance point

The balance point is the position of the jockey at which the meter (galvanometer) shows zero deflection. On either side of it the deflection is in opposite directions.

(c) EMF of the test cell

Given:

• emf of the standard cell, $E_s = 1.50$ V
• balancing length for the standard cell, $l_s = 37.5$ cm
• balancing length for the test cell, $l_t = 25.0$ cm

$$\frac{E_t}{E_s} = \frac{l_t}{l_s} \quad\Rightarrow\quad E_t = 1.50 \times \frac{25.0}{37.5} = 1.00\ \text{V}$$

The emf of the test cell is 1.00 V.

(d) Why the potentiometer is more precise than a voltmeter

• At balance, no current is drawn from the test cell, so there is no $Ir$ drop inside the cell and the true emf is measured. A voltmeter draws current and reads the terminal p.d. $E - Ir$, which is less than $E$.
• It is a null method: it relies only on detecting zero deflection, not on reading a pointer against a scale, so scale and calibration errors of the meter do not matter.
• The result depends on lengths, which can be measured accurately. With a long wire (small potential gradient), even very small differences in emf can be measured.

So the student's argument is correct.