Electrical Circuits

Kirchhoff's law

Kirchhoff's junction rule :
The algebraic sum of the currents into any junction is zero. i.e., \[\Sigma I = 0\] Alternatively, it can also be stated as, 'the sum of currents entering any junction must be equal to the sum of the currents leaving that junction'.

Kirchhoff's loop rule :
The algebraic sum of the potential differences in any loop, including those associated with emf_s and those of resistive elements, must equal zero. \[\Sigma V=0\]

Note that the Kirchhof's junction rule is based on the conservation of electric charge and Kirchhoff's loop rule is based on the conservation of energy.
Sign conventions for the loop rule
In order to apply Kirchhoff's loop rule in electrical circuit, the following sign conventions should be used:
Sign convention for emf_s :
If the direction of travel (i.e., the direction you chose to travel in a loop) is from negative terminal of the cell to positive terminal of the cell, the potential difference across the cell is \(+\mathcal{E}\) where \(\mathcal{E}\) is the emf of the cell. For the direction of travel from positive terminal of the cell to the negative terminal of the cell, the potential difference is \(-\mathcal{E}\). This is shown in figure.

Sign convention for resistances :
On encountering the resistance, if the direction of travel is in the direction of current I, the potential difference across the resistance R is \(-IR\) and if the direction of travel is opposite to the direction of current, the potential difference across the resistance R is \(+IR\). This is shown in figure.

Wheatstone bridge circuit

Wheastone designed a circuit called a 'bridge circuit' which gave an accurate method for measuring the unknown resistance.

If X is the unknown resistance, and P , Q , R are resistance boxes (known resistances). One of these - usually R is adjusted until the galvanometer G between B and D shows no deflection, a so-called 'balance condition'. In this case, the current I_gin G is zero. At balanced condition, \[\frac{P}{Q}=\frac{X}{R}\] Proof:
Wheatstone bridge circuit is shown in figure.
Using Kirchhof's junction rule at junction B,
\[\begin{align*} I_1&=I_3 + I_g\\ I_1&=I_3 \hspace{0.1cm} (\because I_g=0 \hspace{0.1cm} \text{at balanced condition}) \hspace{0.1cm}\text{ ... (i)}\\ \end{align*}\]
At junction D,
\[\begin{align*} I_2 + I_g&=I_4\\ I_2&=I_4 \hspace{0.1cm} (\because I_g=0 \hspace{0.1cm} \text{at balanced condition}) \hspace{0.1cm}\text{ ... (ii)}\\ \end{align*}\]
Using Kirchhoff's loop rule,
In loop ABDA,
\[\begin{align*} -I_1 P -I_g G + I_2 X&=0\\ -I_1 P +I_2 X &=0 \hspace{0.1cm} \because I_g=0\\ I_1 P &= I_2 X \hspace{0.1cm} \text{ ... (iii)}\\ \end{align*}\]
In loop BCDB,
\[\begin{align*} -I_3 Q +I_4 R + I_g G&=0\\ -I_3 Q +I_4 R &=0 \hspace{0.1cm} \because I_g=0\\ I_3 Q &= I_4 R \hspace{0.1cm} \text{ ... (iv)}\\ \end{align*}\]
Dividing (iii) by (iv), we get,
\[\begin{align*} \frac{I_1 P}{I_3 Q}&=\frac{I_2 X}{I_4 R}\\ \therefore \frac{P}{Q}&=\frac{X}{R} \hspace{0.1cm} \text{ ... (v)}\\ \end{align*}\] Equation (v) is the balanced condition of Wheatstone bridge circuit.

Q.1 State and explain Kirchhoff's law of current and voltage. Apply them to obtain the balance condition of Wheatstone bridge. ( See the solution above. )

Meter Bridge

Figure shows the meter bridge / slide wire consisting of 1 m long wire AC. AC has a uniform cross-section and low temperature coefficient of resistance. The unknown resistance X and a known resistance R are connected as in figure. Heavy brass or copper strip is used for the connections AD, FH, JC whose resistances are generally negligible. When the slider is at point C in the wire, it divides the wire into two parts of resistances P and Q as in figure; these, with X and R form a Wheastone bridge.
The 'balance point' is obtained by sliding jockey over the wire AC. Figure shows the Wheatstone bridge equivalent circuit of the meter bridge. At balanced condition, \[\frac{P}{Q}=\frac{R}{X}\hspace{0.1cm} \text{... (i)}\] Let l be the length of wire between A and B, and then (1-l) is the length between B and C (since the total length of AB is 1 m). Since, resistance of a wire is directly proportional to its length,
\[P \propto l \hspace{0.1cm} \text{ ... (ii)}\] and, \[Q \propto (1-l) \hspace{0.1cm} \text {... (iii)}\]
Equation (i) thus becomes, \[\begin{align*} \frac{l}{1-l}&=\frac{R}{X}\\ \therefore X&=\left(\frac{l}{1-l}\right) \times R\\ \end{align*}\]
In this way, the unknown resistance X can be found by using the meter bridge.

Q.2 State the principle of meter bridge. Describe how it can be used to measure the resistance of a wire. ( See the solution above. )

Don't be confused here! Principle of meter bridge is the balanced condition of Wheatstone bridge circuit. So, discuss that briefly.

Potentiometer

Potentiometer is a device which measures the potential difference. It is equivalent to ideal voltmeter (will be discussed in short answer question solution!).
Construction :
The stretched wire potentiometer consists of a long wire AB (as in figure), usually 5 to 10 meters long, fixed on a wooden platform as in figure. The wire has a uniform cross-section. Usually, separate pieces of wire, each 1 m long, are fixed parallel to each other on the platform. The ends A and B are connected to the driving circuit consisting of a strong battery, a plug key and a Rheostat. The driving circuit sends a constant current I through the wire AB.

Working :
One end of a galvanometer is connected to a metal rod fixed on the wooden platform. A 'jockey' may be slid on this metal rod and may touch the wire AB at any desired point P. In this way, the galvanometer gets connected to the point of AB which is touched by the jockey. The length of the wire between the end A and this point can be measured with the help of a meter scale fixed on the platform. The other end C of the galvanometer and the high-potential end A of the wire, form the two end points (terminals) of the potentiometer. These points are connected to the points between which the potential difference is to be measured.
The schematic diagram of the potentiometer is shown in figure.

Principle:
If R be the resistance of the potentiometer wire AB and I be the current flowing through it. Then, V be the potential difference across AP;
\[\begin{align*} V&=IR\\ V&=I\times \frac{\rho l}{A}\\ V&=\frac{I \rho l}{A}\\ \therefore V& \propto l\\ \end{align*}\] Then,
\[V=kl\] where k is the potential gradient i.e., the potential difference per unit length of the wire \(k=\frac{V_0}{l}\)

Determination of internal resistance of a cell

The arrangement of potentiometer circuit for the determination of internal resistance of a cell is shown in figure. The emf of the cell is \(\mathcal{E}\) and its internal resistance is \(r\). A known resistance \(R\) is connected across the battery together with a plug key K. When the key 'K' is open, the balance point P is searched on the wire AB such that there is no deflection in the galvanometer.

As the key K is open, there is no current through the resistance R. Since, there is no current through the battery in this case, and if V be the potential difference across the wire AP, then, \[\mathcal{E}=V\] Since, \[V\propto l\] Then, \[\mathcal{E} \propto l \hspace{0.1cm} \text{ ... (i)}\] Now the key K is closed and the new balanced point P' is searched. The current through the cell is then, \[I=\frac{\mathcal{E}}{R+r}\] Thus, the potential difference across AP' is, \[\begin{align*} V'&=IR\\ &=\frac{\mathcal{E} R}{R+r} \hspace{0.1cm} \text{ ... (ii)}\\ \end{align*}\] Also, if \(l'\) be the new balanced length, then, \[V' \propto l' \hspace{0.1cm} \text{ ... (iii)}\] Dividing equation (iii) by (i), we get, \[\begin{align*} \frac{V'}{\mathcal{E}}&=\frac{l'}{l}\\ \frac{\mathcal{E}R}{\mathcal{E}(R+r)}&=\frac{l'}{l}\\ Rl&=(R+r)l'\\ Rl&=Rl'+rl'\\ \therefore r&=\frac{R(l-l')}{l'}\\ \end{align*}\] In this way, we can determine the internal resistance of the cell.

Comparison of emf_s of two cells

The arrangement of the potentiometer circuit for the comparison of emf_s \(\mathcal{E_1}\) and \(\mathcal{E_2}\) of two cells is shown in figure. A two-way key K₁ and K₂ connects the cells together.

When the key K₁ is plugged, \(\mathcal{E_1}\) is brought into the circuit and lets say that the 'balance point' is obtained at point P₁ (say at length \(l_1\)). So, \[\begin{align*} V_{AP_1}& \propto l_1\\ \therefore \mathcal{E_1}& \propto l_1 \hspace{0.1cm} \text{ ... (i)}\\ \end{align*}\]
When the key K₂ is plugged , \(\mathcal{E_2}\) is brought into the circuit and lets say that the 'balance point' is obtained at point P₂ (say at length \(l_2\)). So, \[\begin{align*} V_{AP_2}& \propto l_2\\ \therefore \mathcal{E_2}& \propto l_2 \hspace{0.1cm} \text{ ... (ii)}\\ \end{align*}\]
Dividing equation (ii) by (i), we get,
\[\begin{align*} \frac{\mathcal{E_1}}{\mathcal{E_2}}&=\frac{l_1}{l_2}\\ \end{align*}\]
In this way, the emf_s of two cells can be compared.

Q.3 State the principle of potentiometer. Describe how it is used to determine the internal resistance of a cell. ( See the solution above. )

Q.4 State the principle of potentiometer. Describe how it is used to compare the internal resistance of two cells. ( See the solution above. )

Superconductors

Superconductors are the materials with zero electrical resistance. For example, Nb₃Ge, YBa₂Cu₃O₇, etc.
A short background :
In 1911, Kamerlingh Onnes while observing the electrical resistance of mercury at very low temperatures close to 4.2 K observed that the electrical resistance of mercury decreased continuously from its melting point (233 K) to 4.2 K and then within some hundreths of a degree, dropped suddenly to about a millionth of its original value at the melting point as shown in figure. The phenomenon of disappearance of electrical resistance of material below a critical temperature was called superconductivity by Onnes and the material in this state was called superconductors.

Perfect conductors

A perfect conductor is an electrical conductor with no resistivity.

Difference between perfect conductors and superconductors:

Superconductors are obtained in real life, while perfect conductors are the assumption to model system in which electrical resistance/resistivity is negligible compared to other effects.
Perfect conductors can have any temperature, but superconductors only exist below the critical temperature (in the case of mercury, the critical temperature is 4.2 K).

Galvanometer

A galvanometer is a device that detects the small amount of current flowing through it. It can be converted into ammeter and voltmeter by connecting a suitable resistor appropriately.
For the conversion of galvanometer into ammeter, a small resistance 'Shunt' is used in parallel with the galvanometer, and a very high resistance is used in series with the galvanometer to convert it into a voltmeter.

Conversion of galvanometer into ammeter

An galvanometer can be converted into ammeter by using shunt in parallel to it. Let G be the resistance of the galvanometer, S be the shunt and I be the maximum current to be measured by an ammeter. I_g be the maximum current flowing through the galvanometer and I_s be the current through shunt.

Since the galvanometer and shunt are in parallel to each other, the potential difference across them must be equal. p.d. across G = p.d. across S \[\begin{align*} I_g G&=I_s S\\ I_g G&=(I-I_g)S\\ \therefore S&=\frac{I_g G}{I-I_g}\\ \end{align*}\] So this is the value of shunt that must be connected in parallel to the galvanometer to convert it to an ammeter.
The resistance of the resulting ammeter is then the parallel combination of resistances of galvanometer and shunt. So, the equivalent resistance (i.e., resistance of ammeter R_A ) is, \[\begin{align*} \frac{1}{R_A}&=\frac{1}{G}+\frac{1}{S}\\ \frac{1}{R_A}&=\frac{S+G}{GS}\\ \therefore R_A &=\frac{GS}{G+S}\\ \end{align*}\]

Conversion of galvanometer into voltmeter

Let G be the resistance of the galvanometer and I_g be the maximum current flowing through the galvanometer. R be the resistance connected in a series with a galvanometer and V be the maximum voltage measured by the voltmeter.

Since, R and G are in series, the total potential difference (measured by the voltmeter) must be equal to the sum of potential differences across R and V.
V = p.d. across G + p.d. across R \[\begin{align*} V&=I_gG+I_gR\\ I_gR&=V=I_gG\\ \therefore R&=\frac{V}{I_g}-G\\ \end{align*}\] Hence, this is the value of resistance that must be connected in series to the galvanometer to convert it to an voltmeter.
The resistance of the resulting voltmeter is then the series combination of resistances of galvanometer and R. So, the equivalent resistance (i.e., resistance of voltmeter R_A ) is, \[R_v=G+R\]

Q.5 Define Shunt. How can you convert galvanometer into ammeter? ( See the solution above. )

Q.6 Define galvanometer. How can you convert galvanometer into voltmeter? ( See the solution above. )

Short answer question solution

Q.1 You are asked to measure the emf of a cell. Which instrument will you use to measure: a high resistance voltmeter or a potentiometer and why?
Solution:
In a voltmeter, a resistor having a high resistance R is connected in series with the galvanometer having resistance G. When the voltmeter is used in a circuit, its resistance \[R_v=R+G\] is connected in parallel to some element in the circuit. This changes the overall current in the circuit and hence affects the potential difference across that element. Due to this reason, though having large resistance, some current is drawn by the voltmeter. However, potentiometer draws no current through the circuit and still measures the potential difference. So, I would use potentiometer to measure the given emf of a cell.
Q.2 While measuring the emf of a cell, the circuit should be open. How is it achieved in potentiometer?
Solution:
For an open circuit, the current should not flow through the cell. In a potentiometer, a 'balancing length' is obtained in the potentiometer wire for which no current flows through the galvanometer and hence through the circuit consisting of a cell. Then, the potential difference across the balancing length \(l\) is measured by using the relation, \[V=kl\] where, k is the potential gradient. This potential difference is equal to the emf of a cell. In this way, the emf of a cell is measured by making the circuit open as desired.
Q.3 Why do we prefer a potentiometer with a longer bridge wire?
Solution:
The maximum potential difference that can be measured by the potentiometer depends on the length of the potentiometer because for potentiometer, \[V \propto l\] where \(l\) is the length of the potentiometer wire. More the length of the potentiometer wire, more is the potential difference is obtained and thus larger potential difference can be measured across the desired elements. Hence, longer potentiometer wire is preferred.
Q.4 Using meter bridge, it is advised to obtain the null point in the middle of the wire. Why?
Solution:
The variable resistance R in a meter bridge should be chosen in such a way that the balance point comes fairly near to the middle of the wire. If either of the length of the arms is small, the resistance of its end connection (i.e., resistance of copper strips) is not negligible in comparison with its own resistance. Hence, the balanced condition equation will not hold in this case. Thus, it is advised to obtain the null point in the middle of the wire in meter bridge.
Q.5 An ammeter is always connected in series. Why?
Solution:
Same current flows through the circuit in a series. Thus to measure the current through a specific circuit, it should be connected in a series to that circuit. Another reason is that, it has a very low resistance and draws negligible amount of current through the circuit.
Q.6 Voltmeter is always connected in parallel. Why?
Solution:
Same potential difference is obtained across the elements in parallel. Thus to measure the p.d. across the element, it should be connected in parallel to the circuit. Another reason is that, it has very high resistance and draws very small current which doesnot alter the p.d. across the terminals considerably.
Q.6 Potentiometer is an ideal voltmeter. Explain.
Solution:
Refer to q.no. (1) .

Numerical Problem solution

Q. 1 What must be the emf E in the circuit so that the current flowing through the 7 Ω resistor is 1.80 A? Each emf source has negligible internal resistance. (Click here for the solution!)

Q. 2 Using Kirchhoff's laws of current and voltage, find the current in 2 Ω resistor in the given circuit. (Click here for the solution!)

Q. 3 In the adjacent circuit find:
(i) the current in resistor R,
(ii) resistance R,
(iii) the unknown emf E,
(iv) if the circuit is broken at P, what is the current in resistor R? (Click here for the solution!)

Q. 4 A potentiometer is 10 m long. It has a resistance of 20 Ω. It is connected in series with a battery of 3 V and a resistance of 10 Ω. What is the potential gradient along with wire? (Click here for the solution!)

Q. 5 A simple potentiometer circuit is setup as in fig., using uniform wire AB, 1.0 m long, which has a resistance of 2 Ω . The resistance of the 4 V battery is negligible. If the variable resistor R were given a value of 2.4 Ω, what would be the length AC for Zero galvanometer deflection? (Click here for the solution!)

Q. 6 The total length of the wire of a potentiometer is 1.0 m. A potential gradient of 0.0015 V/cm is obtained when a steady current is passed through this wire. Calculate,
(i) the distance of the null point on connecting standard cell of 1.018 V.
(ii) the unknown p.d. if the null point is obtained ata distance of 940 cm, and (iii) the maximum p.d. which can be measured by this instrument. (Click here for the solution!)

Q. 7 The driver cell of a potentiometer has an emf of 2 V and negligible internal resistance. The potentiometer wire has a resistance of 3 Ω. Calculate the resistance needed in series with a wire if a p.d. 5.0 mV is required across the whole wire. The wire is 100 cm long and a balanced length of 60 cm is obtained for a thermocouple of emf E. What is the value of E? (Click here for the solution!)

Q. 8 The emf of a battery A is balanced by a length 75 cm on a potentiometer wire. The emf of a standard cell 1.02 V is balanced by a length of 50 cm. What is the emf of A? (Click here for the solution!)

Physics in Depth