Electrical Circuits
Class 12 Physics Notes • NEB Syllabus 2083
Syllabus, 14 solved multiple choice questions, step-by-step numericals and derivations, handwritten and typed PDF notes and frequently asked questions on Kirchhoff’s laws, bridges, the potentiometer, meters and Joule’s law.
1. Syllabus
Unit: Electricity and Magnetism
Chapter 14: Electrical Circuits (8 hours)
These notes follow the NEB Class 12 Physics syllabus (+2 Physics, Nepal). Practise the multiple choice questions first, then work through the numericals, and finish with the PDF notes for quick revision.
2. Multiple Choice Questions
Question 1. Kirchhoff’s junction law is based on the conservation of:
- a) energy
- b) charge
- c) momentum
- d) mass
Correct Answer: b) charge
Reason: The total charge entering a junction per second must equal the total charge leaving it, because charge can neither accumulate at a junction nor be destroyed. Hence \(\sum I=0\) at a junction.
Question 2. Kirchhoff’s loop law is a statement of the conservation of:
- a) charge
- b) energy
- c) momentum
- d) current
Correct Answer: b) energy
Reason: When a unit charge goes around a closed loop and returns to its starting point, its net gain of energy is zero. The emfs supply the energy and the resistors dissipate it, so
Question 3. The condition for a balanced Wheatstone bridge with resistances \(P,Q\) (one arm) and \(R,S\) (the other arm) is:
- a) \(\dfrac PQ=\dfrac RS\)
- b) \(\dfrac PQ=\dfrac SR\)
- c) \(PQ=RS\)
- d) \(P+Q=R+S\)
Correct Answer: a) \(\dfrac PQ=\dfrac RS\)
Reason: When the galvanometer shows no deflection, the points B and D are at the same potential, and Kirchhoff’s laws give
Question 4. In a balanced Wheatstone bridge \(P=10\ \Omega,\ Q=20\ \Omega\) and \(R=15\ \Omega\). The resistance \(S\) is:
- a) \(7.5\ \Omega\)
- b) \(15\ \Omega\)
- c) \(30\ \Omega\)
- d) \(45\ \Omega\)
Correct Answer: c) \(30\ \Omega\)
Reason: From \(\dfrac PQ=\dfrac RS\):
Question 5. In a meter bridge, a resistance \(R=10\ \Omega\) is in the left gap and the null point is at \(40\ \mathrm{cm}\) from the left end. The unknown resistance in the right gap is:
- a) \(6.7\ \Omega\)
- b) \(10\ \Omega\)
- c) \(15\ \Omega\)
- d) \(25\ \Omega\)
Correct Answer: c) \(15\ \Omega\)
Reason:
Question 6. A potentiometer is preferred to a voltmeter for measuring the emf of a cell because:
- a) it is cheaper than a voltmeter
- b) it draws no current from the cell at the balance point
- c) it has a very low resistance
- d) it has a very short wire
Correct Answer: b) it draws no current from the cell at the balance point
Reason: At the null point no current flows through the cell under test, so the potentiometer measures the true emf. A voltmeter draws some current, which causes a drop across the internal resistance, and reads less than the emf.
Question 7. Two cells balance at \(60\ \mathrm{cm}\) and \(40\ \mathrm{cm}\) on a potentiometer wire. The ratio of their emfs \(\varepsilon_1:\varepsilon_2\) is:
- a) 2 : 3
- b) 3 : 2
- c) 9 : 4
- d) 1 : 1
Correct Answer: b) 3 : 2
Reason: The emf is proportional to the balance length:
Question 8. The internal resistance of a cell, found with a potentiometer, is (here \(l_1\) is the balance length on open circuit and \(l_2\) the balance length when a resistance \(R\) is connected across the cell):
- a) \(r=R\,\dfrac{l_1-l_2}{l_2}\)
- b) \(r=R\,\dfrac{l_2}{l_1-l_2}\)
- c) \(r=R\,\dfrac{l_1}{l_2}\)
- d) \(r=R\,\dfrac{l_1+l_2}{l_2}\)
Correct Answer: a) \(r=R\,\dfrac{l_1-l_2}{l_2}\)
Reason: \(\dfrac{\varepsilon}{V}=\dfrac{R+r}{R}=\dfrac{l_1}{l_2}\), so
Question 9. The resistance of a superconductor becomes zero below a particular temperature called the:
- a) Curie temperature
- b) critical temperature
- c) boiling point
- d) room temperature
Correct Answer: b) critical temperature
Reason: Every superconductor has a critical (transition) temperature \(T_c\). Below \(T_c\) its resistance drops to zero. For example, mercury becomes superconducting at about \(4.2\ \mathrm{K}\).
Question 10. Which property distinguishes a superconductor from a merely perfect conductor?
- a) Zero electrical resistance
- b) Expulsion of the magnetic field from its interior (Meissner effect)
- c) A very high melting point
- d) A very high thermal conductivity
Correct Answer: b) Expulsion of the magnetic field from its interior (Meissner effect)
Reason: Both have zero resistance. A perfect conductor keeps the magnetic flux that is already inside it, but a superconductor expels the magnetic field from its interior when it is cooled below \(T_c\) and behaves as a perfect diamagnet.
Question 11. A galvanometer is converted into an ammeter by connecting:
- a) a high resistance in series
- b) a high resistance in parallel
- c) a low resistance (shunt) in parallel
- d) a low resistance in series
Correct Answer: c) a low resistance (shunt) in parallel
Reason: The shunt carries most of the current, so only the safe current \(I_g\) passes through the galvanometer. The value of the shunt is
Question 12. A galvanometer is converted into a voltmeter by connecting:
- a) a high resistance in series
- b) a high resistance in parallel
- c) a low resistance in series
- d) a low resistance in parallel
Correct Answer: a) a high resistance in series
Reason: The series resistance limits the current through the galvanometer to \(I_g\) for the full range \(V\). Its value is
Question 13. According to Joule’s law, if the current through a resistor is doubled (the resistance and the time remaining the same), the heat produced becomes:
- a) two times
- b) four times
- c) half
- d) eight times
Correct Answer: b) four times
Reason: \(H=I^2Rt\), so \(H\propto I^2\):
Question 14. In a series ohmmeter the pointer reads zero ohm when:
- a) the terminals are shorted, giving full-scale deflection
- b) the terminals are open, giving no deflection
- c) a very high resistance is connected
- d) the cell is removed
Correct Answer: a) the terminals are shorted, giving full-scale deflection
Reason: When the terminals are joined, the resistance between them is zero and the current is maximum, giving full-scale deflection. This position is marked zero, and an open circuit (infinite resistance) is marked at the other end, so the scale is read from right to left.
3. Numerical and Subjective Questions
Figure: Circuit for the application of Kirchhoff’s laws
a) State Kirchhoff’s current law and Kirchhoff’s voltage law.
Kirchhoff’s current law (junction rule): the algebraic sum of the currents meeting at a junction is zero. It is based on the conservation of charge.
Kirchhoff’s voltage law (loop rule): the algebraic sum of all the changes of potential (emfs and \(IR\) drops, taken with the signs given below) around any closed loop of a circuit is zero. It is based on the conservation of energy.
b) State the sign conventions used when the loop law is applied.
c) Find the currents \(I_1,I_2\) and \(I_3\) in the circuit shown in the figure. Neglect the internal resistances of the cells.
Junction rule at B:
Loop 1 (D→A→B→E→D):
Loop 2 (F→C→B→E→F):
Subtracting (2) from (1):
The negative sign of \(I_2\) shows that the actual current in the branch of \(E_2\) flows opposite to the assumed direction, so the cell \(E_2\) is being charged by \(E_1\).
Check (loop 1): \(12-2(2)-6\left(\dfrac43\right)=12-4-8=0\).
Figure: Wheatstone bridge (the bridge is balanced when the galvanometer shows no deflection)
a) What is a Wheatstone bridge? State its balance condition.
A Wheatstone bridge consists of four resistances \(P,Q,R\) and \(S\) connected in the form of a quadrilateral ABCD. A cell is connected across one diagonal (A and C) and a galvanometer across the other diagonal (B and D). The bridge is said to be balanced when the galvanometer shows no deflection. The balance condition is
b) Derive the balance condition of a Wheatstone bridge using Kirchhoff’s laws.
Let the current \(I_1\) flow through \(P\) and \(I_2\) through \(R\). At balance no current flows through the galvanometer (\(I_g=0\)), so the same current \(I_1\) flows through \(Q\) and \(I_2\) through \(S\).
Applying the loop law to the loop ABDA (the galvanometer drop is zero):
Applying the loop law to the loop BCDB:
Dividing (1) by (2):
c) In a balanced Wheatstone bridge \(P=5\ \Omega,\ Q=10\ \Omega\) and \(R=12\ \Omega\). Calculate the unknown resistance \(S\).
d) What happens to the balance condition if the cell and the galvanometer are interchanged?
The balance condition does not change. The condition \(PS=QR\) is symmetrical, so the bridge remains balanced when the positions of the cell and the galvanometer are interchanged.
Figure: Meter bridge
a) Describe the principle of a meter bridge.
A meter bridge works on the principle of the Wheatstone bridge. The two arms \(P\) and \(Q\) are replaced by the two parts \(AJ\) and \(JC\) of a uniform wire, whose resistances are directly proportional to their lengths. The known resistance \(R\) and the unknown resistance \(S\) are connected in the two gaps, and the jockey \(J\) is moved on the wire until the galvanometer shows no deflection.
b) Derive the expression for the unknown resistance with a meter bridge.
Let the balance point \(J\) be at a distance \(l\) cm from \(A\). If \(\sigma\) is the resistance of the wire per cm, then
At balance the condition of the Wheatstone bridge gives \(\dfrac{P}{Q}=\dfrac RS\), hence
c) A resistance of \(4\ \Omega\) is connected in the left gap of a meter bridge, and the null point is obtained at \(40\ \mathrm{cm}\) from the left end. Calculate the unknown resistance in the right gap.
d) Why is the meter bridge most accurate when the null point is near the middle of the wire?
The percentage error in \(S\) depends on the error in measuring \(l\) and is least when \(l\) is near \(50\ \mathrm{cm}\), because the sum \(l+(100-l)\) is fixed and the relative error in both \(l\) and \(100-l\) is then smallest. For this reason \(R\) is chosen to be of the same order as \(S\).
Figure: Potentiometer circuit (key K open for emf; key K closed for internal resistance)
a) State the principle of a potentiometer.
A potentiometer consists of a long uniform wire AB through which a steady current is passed from a driver cell. The potential drop across any length of the wire is directly proportional to that length:
where \(k=\dfrac{I\rho}{A}\) is the potential gradient, the fall of potential per unit length of the wire.
b) Describe how the emfs of two cells can be compared with a potentiometer.
The positive terminal of the cell under test is connected to A, and its negative terminal is connected through a galvanometer to the jockey \(J\). The jockey is moved until the galvanometer shows no deflection, and the balance length \(l_1\) is measured for the cell \(\varepsilon_1\). The experiment is repeated for the second cell \(\varepsilon_2\) to find the balance length \(l_2\). At the null point no current flows through the cell, so its emf equals the potential drop across the balance length:
c) A standard cell of emf \(1.08\ \mathrm{V}\) balances at \(54\ \mathrm{cm}\) of a potentiometer wire, and an unknown cell balances at \(72\ \mathrm{cm}\). Find the emf of the unknown cell.
d) Why is a potentiometer better than a voltmeter for measuring the emf of a cell?
At the balance point a potentiometer draws no current from the cell, so there is no potential drop across the internal resistance and the potentiometer measures the true emf. A voltmeter draws some current and therefore reads only the terminal potential difference, which is less than the emf. A potentiometer also acts as a voltmeter of infinite resistance.
a) Derive the expression for the internal resistance of a cell using a potentiometer.
Key K open (open circuit): the balance length is \(l_1\), and the emf is
Key K closed: a known resistance \(R\) is connected across the cell. If the balance length is \(l_2\), the terminal potential difference \(V\) is
The current through the cell is \(I=\dfrac{\varepsilon}{R+r}\), so \(V=IR=\dfrac{\varepsilon R}{R+r}\), and
b) A cell has a balance length of \(75\ \mathrm{cm}\) on open circuit. When a resistance of \(10\ \Omega\) is connected across it, the balance length becomes \(60\ \mathrm{cm}\). Calculate the internal resistance of the cell.
c) Why is a potentiometer used instead of a voltmeter to find the internal resistance?
A voltmeter draws a current from the cell, which changes the terminal voltage, and it cannot read the emf. The potentiometer reads both the emf (open circuit) and the terminal voltage (closed circuit) accurately without drawing any current at the balance point.
a) How can the sensitivity of a potentiometer be increased?
The sensitivity is increased by decreasing the potential gradient \(k\), because a small difference of potential then corresponds to a large change in the balance length. This is done by (i) using a longer potentiometer wire and (ii) reducing the current in the wire with a rheostat.
b) A potentiometer wire of length \(10\ \mathrm{m}\) and resistance \(20\ \Omega\) is connected in series with a battery of emf \(3\ \mathrm{V}\) and a rheostat of resistance \(10\ \Omega\). Neglect the internal resistance of the battery. Find (i) the potential gradient and (ii) the balance length for a cell of emf \(1.2\ \mathrm{V}\).
b) i) The potential gradient.
b) ii) The balance length for the cell of emf \(1.2\ \mathrm{V}\).
c) Why must the emf of the driver cell be greater than the emf of the cell to be measured?
The potential drop across the whole wire AB must be at least equal to the emf being balanced. If the driver cell had a smaller emf, no point on the wire would have a potential equal to the emf of the test cell, and no balance point would be found.
a) What is a superconductor? What is the critical temperature?
A superconductor is a material whose electrical resistance becomes exactly zero when it is cooled below a certain temperature. This temperature is called the critical temperature \(T_c\). For example, the resistance of mercury drops abruptly to zero at about \(4.2\ \mathrm{K}\).
b) Differentiate between a superconductor and a perfect conductor.
| Perfect conductor | Superconductor | |
|---|---|---|
| Resistance | Zero at all temperatures (an ideal conductor). | Zero only below the critical temperature \(T_c\). |
| Magnetic field inside | The field already present is trapped; the field inside stays unchanged. | The magnetic field is expelled and \(B=0\) inside (Meissner effect). |
| Magnetic behaviour | Depends on how the material was prepared. | A perfect diamagnet, independent of the history of the material. |
c) Mention some applications of superconductors.
Figure: Conversion of a galvanometer into an ammeter (left) and into a voltmeter (right)
a) Why cannot a galvanometer be used directly to measure a large current?
A galvanometer is a very sensitive device. A large current would give a deflection beyond the scale and would damage its coil. It can safely carry only a small full-scale current \(I_g\) (of the order of milliamperes).
b) Show how a galvanometer is converted into an ammeter of a given range. Derive the expression for the shunt resistance.
A small resistance \(S\), the shunt, is connected in parallel with the galvanometer of resistance \(G\). Let the ammeter be required to read a maximum current \(I\). Of this, \(I_g\) passes through the galvanometer and \(I-I_g\) through the shunt. The two are in parallel and have the same potential difference:
The effective resistance of the ammeter is \(R_A=\dfrac{GS}{G+S}\), which is very small.
c) A galvanometer of resistance \(100\ \Omega\) gives full-scale deflection for a current of \(1\ \mathrm{mA}\). Calculate the shunt required to convert it into an ammeter of range \(0\)–\(1\ \mathrm{A}\), and the resistance of the ammeter.
d) Why should an ideal ammeter have zero resistance?
An ammeter is connected in series. If its resistance is not zero, it adds to the resistance of the circuit and reduces the current that it is meant to measure. An ideal ammeter has zero resistance so that it does not change the current.
a) Show how a galvanometer is converted into a voltmeter. Derive the expression for the series resistance.
A high resistance \(R\) is connected in series with the galvanometer of resistance \(G\). For the voltmeter to read a maximum potential difference \(V\), the current \(I_g\) must flow through the combination:
The resistance of the voltmeter, \(G+R\), is very high. The voltmeter is connected in parallel with the component across which the potential difference is to be measured.
b) A galvanometer of resistance \(50\ \Omega\) shows full-scale deflection for \(2\ \mathrm{mA}\). Calculate the resistance to be connected in series to convert it into a voltmeter of range \(0\)–\(10\ \mathrm{V}\).
c) What additional resistance must be connected in series with this voltmeter to extend its range to \(20\ \mathrm{V}\)?
d) Why should an ideal voltmeter have infinite resistance?
A voltmeter is connected in parallel. If its resistance is not very high, it draws a current from the circuit and changes the potential difference that it is meant to measure. An ideal voltmeter has infinite resistance so that it draws no current.
Figure: Series ohmmeter
a) Describe the principle and working of an ohmmeter.
A series ohmmeter consists of a cell of emf \(E\), a galvanometer and a variable resistance \(R\) connected in series with two terminals. The unknown resistance \(R_x\) is connected across the terminals. If \(R_0\) is the total resistance of the cell, galvanometer and \(R\), the current is
When the terminals are shorted (\(R_x=0\)) the current is maximum, and \(R\) is adjusted to give full-scale deflection, which is marked \(0\ \Omega\). When the terminals are open (\(R_x=\infty\)) there is no deflection, marked \(\infty\). The scale is therefore read from right to left, and it is not uniform because \(I\) is not proportional to \(R_x\).
b) An ohmmeter uses a cell of emf \(1.5\ \mathrm{V}\) and a galvanometer of full-scale current \(1\ \mathrm{mA}\). The variable resistance is adjusted so that the full-scale deflection occurs when the terminals are shorted. Find:
b) i) The total resistance \(R_0\) of the ohmmeter.
b) ii) The unknown resistance that gives a half-scale deflection.
b) iii) The current when \(R_x=500\ \Omega\).
This is three-fourths of the full-scale deflection.
a) State Joule’s law of heating.
Joule’s law: the heat produced in a conductor is directly proportional to (i) the square of the current, (ii) the resistance of the conductor and (iii) the time for which the current flows.
b) Derive the expression for the heat produced in a conductor.
When a charge \(q\) moves through a potential difference \(V\) across a resistor, the work done (the electrical energy converted into heat) is \(W=qV\). Since \(q=It\):
Using Ohm’s law \(V=IR\):
c) A current of \(2\ \mathrm{A}\) flows through a resistor of \(25\ \Omega\) for \(10\) minutes. Calculate the heat produced in joules and in calories (\(1\ \mathrm{cal}=4.18\ \mathrm{J}\)).
d) An electric heater is rated \(1100\ \mathrm{W}\) at \(220\ \mathrm{V}\). Calculate its resistance, the current drawn and the heat produced in \(2\) minutes.
a) Mention some applications of the heating effect of current.
b) Why is nichrome used as the heating element and tungsten as the filament of a bulb?
The heating element is made of nichrome because it has a high resistivity, a high melting point and does not oxidise easily when it is red hot. Tungsten is used for the bulb filament because it has a very high melting point (about \(3400^\circ\mathrm{C}\)) and can therefore be heated to white heat without melting.
c) Why is the fuse wire made of a material of low melting point and high resistance?
The fuse wire must melt quickly when the current becomes larger than a safe value, so that the circuit is broken and the appliances are protected. A low melting point and a high resistance make \(H=I^2Rt\) raise its temperature to the melting point rapidly.
d) An electric kettle is rated \(2\ \mathrm{kW}\). How long does it take to heat \(1\ \mathrm{kg}\) of water from \(20^\circ\mathrm{C}\) to \(100^\circ\mathrm{C}\)? Assume that all the heat is absorbed by the water (specific heat capacity of water \(=4200\ \mathrm{J\,kg^{-1}\,K^{-1}}\)).
4. View PDF Notes
i. Handwritten Notes
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5. Frequently Asked Questions
What is an electrical circuit?
What topics are included in Chapter 14 Electrical Circuits of NEB Class 12 Physics?
What are Kirchhoff’s laws?
What is the balance condition of a Wheatstone bridge?
What is the principle and formula of a meter bridge?
What is the principle of a potentiometer and why is it better than a voltmeter?
How do you compare the emfs of two cells using a potentiometer?
How do you measure the internal resistance of a cell using a potentiometer?
What is the difference between a superconductor and a perfect conductor?
How is a galvanometer converted into an ammeter?
How is a galvanometer converted into a voltmeter?
How does an ohmmeter work?
What is Joule’s law of heating?
What are the main components in an electric circuit diagram?
What are the types of electrical circuits?
Is this chapter the same as the Electrical Circuits 1 book or PDF?
Where can I get Class 12 Physics Chapter 14 Electrical Circuits notes in PDF?
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