WBCHSE Class 12 Physics Kirchhoff’s Laws And Electrical Measurement Multiple Choice Questions

WBCHSE Class 12 Physics MCQs

Kirchhoff’s Laws And Electrical Measurement Multiple Choice Question And Answers

Question 1. Kirchhoff’s laws are valid for

1. Linear circuits only
2. Non-linear circuits only
3. Both linear and non-linear circuits
4. None of the above

Answer: 3. Both linear and non-linear circuits

Question 2. For the following circuit, the voltage across AB is

1. 2V
2. 46V
3. 3V
4. 0.2V

Answer: 1. 2V

Question 3. In the circuit, the current I₂ exceeds the current I₁, by a factor of

Read and Learn More Class 12 Physics Multiple Choice Questions

1. 12
2. 20
3. 100
4. 120

Answer: 4. 120

WBBSE Class 12 Kirchhoff’s Laws MCQs

Question 4. When the switch S is closed the current passing through the 4Ω resistance is

1. 4.5A
2. 6A
3. 3A
4. zero

Answer: 1. 4.5A

Question 5. In the circuit, if the potential at point A is taken to be zero, the potential at point B is

1. +1V
2. -IV
3. +2V
4. -2 V

Answer: 1. +1V

Question 6. In the given network, if the potential at point A is taken to be zero, the potential at point B is

1. -IV
2. 2V
3. -2V
4. IV

Answer: 4. IV

Question 7. In the given network the magnitude of currents is shown here. The current I will be

1. -3 A
2. 3A
3. 13 A
4. 23 A

Answer: 3. 13 A

WBCHSE class 12 physics MCQs

Question 8. Current through 2Ω resistance is

1. Zero
2. 2 A
3. 4A
4. 5A

Answer: 1. Zero

Key Concepts in Kirchhoff’s Laws

Question 9. Current through 2Ω resistance is

1. 0
2. 2A
3. 4A
4. 5A

Answer: 1. 0

Question 10. In the circuit, the source of emf E has negligile internal resistance. C is the midpoint of the potentiometer wire AB. The resistance of the voltmeter V is not very high compared to that of the potentiometer wire. Then the voltmeter reading will be

1. E
2. \(\frac{E}{2}\)
3. greater than \(\frac{E}{2}\)
4. less than \(\frac{E}{2}\)

Answer: 4. less than \(\frac{E}{2}\)

Question 11. For a potentiometer wire of fixed length, the potential gradient can be decreased by

1. Increasing the current by the potentiometer wire
2. Reducing the current in the potentiometer wire
3. Decreasing the value of attached resistances
4. None of the above

Answer: 2. Reducing the current in the potentiometer wire

Question 12. In which case, will the null condition of a Wheatstone bridge change?

1. If the resistances in different arms are changed
2. If the positions of the battery and the galvanometer are
   interchanged
3. If a battery of different emf is used
4. If a galvanometer of different resistance is used

Answer: 1. If the resistances in different arms are changed

Question 13. The resistances of the four arms of a Wheatstone bridge are 1Ω, 3Ω, 2Ω and 6Ω respectively and the resistance of the galvanometer is 1000Ω. The equivalent resistance of the combination is

1. 12Ω
2. 1000Ω
3. 2.67Ω
4. 2.4Ω

Answer: 2.67Ω

Question 14. If the value of each resistance is R then what will be the equivalent resistance between the points A and B?

1. \(\frac{R}{5}\)
2. \(\frac{R}{2}\)
3. R
4. \(\frac{3R}{2}\)

Answer: 3. R

Question 15. The resistances in the first and the second arms of a Wheatstone bridge are P = 10Ω and Q = 20Ω. The third and the fourth arm resistances R and S are so chosen that the bridge is balanced. Now, R is kept fixed, but P and Q are interchanged. The new value of the fourth arm resistance at balance is x% of the old value of S. The value of x is

1. 25
2. 50
3. 200
4. 400

Answer: 1. 25

Short Answer Questions on Kirchhoff’s Current Law

Question 16. Five equal resistances, each of resistance R, are connected. A battery of V volt is connected between A and B. The current flowing in FC will be

1. \(\frac{3V}{R}\)
2. \(\frac{V}{R}\)
3. \(\frac{V}{2R}\)
4. \(\frac{2V}{R}\)

Answer: 3. \(\frac{V}{2R}\)

Question 17. It is observed that the current is independent of the value of the resistance R6. Then the resistance values must satisfy the relation

1. R₁R₂R₃ = R₃R₄R₅
2. R₁R₄ = R₂R₃
3. \(\frac{1}{R_5}+\frac{1}{R_6}=\frac{1}{R_1+R_2}+\frac{1}{R_3+R_4}\)
4. R₁R₃ = R₂R₄ = R₅R₆

Answer: 2. R₁R₄ = R₂R₃

Question 18. Copper wire is not used as the bridge wire in a meter bridge because

1. Resistance of copper wire changes due to changes in temperature
2. The resistance of a copper wire is very small
3. In the case of copper, the error due to the thermoelectric effect is very great
4. The thermoelectric effect sets in at the two ends of a copper

Answer: 2. Resistance of a copper wire is very small

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Question 19. The effective length of the wire of a metre bridge is, in general, more than 1 m because of

1. Joule heating of the wire
2. Thermoelectric effect at the two ends of the wire
3. Junction defects at the two ends of the wire
4. Elastic stress generated in the wire

Answer: 3. Junction defects at the two ends of the wire

Question 20. A resistance of 1Ω is kept in the left gap of a metre bridge and another resistance of 3Ω is kept in the right gap. The left and right end errors of the metre wire are 3 cm and 1 cm respectively. The null point would be at

1. 22.0 cm
2. 23.0 cm
3. 25.0 cm
4. 26.0 cm

Answer: 2. 23.0 cm

Question 21. Resistance in the two gaps of a meter bridge are 10Ω and 30Ω respectively. If the resistances are interchanged the balance point shifts by

1. 33.3 cm
2. 66.67 cm
3. 25 cm
4. 50 cm

Answer: 4. 50cm

Question 22. A student chooses the standard resistance S to be 100 XI while measuring a resistance R by using a metre bridge. He finds the null point at l₁ = 2.9 cm. He is told to attempt to improve accuracy. Which of the following is a useful way?

1. He should measure l₁ more accurately
2. He should change S to 1000Ω and repeat the experiment
3. He should change S to 3Ω and repeat the experiment
4. He should give up hope of a more accurate measurement with a metre bridge

Answer: 3. He should change S to 3Ω and repeat the experiment

⇒ \(R=S \times \frac{l_1}{100-l_1}=\frac{100 \times 2.9}{100-2.9}=3 \Omega\)

Question 23. Two cells of emf’s approximately 5 V and 10 V are to be accurately compared using a potentiometer of length 400 cm.

1. The battery that runs the potentiometer should have a voltage of 8 V
2. The battery of the potentiometer can have a voltage of 15 V and R adjusted so that the potential drop across the wire slightly exceeds 10 V
3. The first 50 cm portion of the wire itself should have a potential drop of 10 V
4. The potentiometer is usually used for comparing resistances and not voltages

Answer: 2. The battery of the potentiometer can have a voltage of 15 V and R adjusted so that the potential drop across the wire slightly exceeds 10 V

Question 24. Kirchhoff’s junction rule is a reflection of

1. Conservation Of Current Density Vector
2. Conservation Of Charge
3. The fact that the momentum with which a charged particle approaches a junction is unchanged (as a vector) as the charged particle leaves the junction
4. The fact there is no accumulation of charges at the junction

Answer:

2. Conservation Of Charge

4. The fact there is no accumulation of charges at the junction

WBCHSE class 12 physics MCQs

Question 25. The measurement of an unknown resistance R₄ is to be carried out using outusingWheatstonebridge. Two students perform an experiment in two ways. The first student takes R₂ = 10Ω and R₁= 5Ω. The other student takes R₂ = 1000Ω and R₁ = 500Ω. In the standard arm, both take R₃ = 5Ω. Both find \(R_4=\frac{R_2}{R_1} \times R_3=10 \Omega\) within errors.

1. The errors of measurement of the two students are the same
2. Errors of measurement do depend on the accuracy with which R₂ and R₁ can be measured
3. If the student uses large values of R₂ and R₁, the currents through the arms will be feeble. This will make a determination of a null point more difficult
4. Wheatstone bridge is a very accurate instrument and has no errors in measurement

Answer:

2. Errors of measurement do depend on the accuracy with which R₂ and R₁ can be measured

3. If the student uses large values of R₂ and R₁, the currents through the arms will be feeble. This will make a determination of a null point more difficult

Common MCQs on Kirchhoff’s Voltage Law

Question 26. Consider a simple circuit. JR’ is a variable resistance which can vary from R₀ to infinity, r is the internal resistance of the battery (r<< R << R₀)

1. The potential drop across AB is nearly constant as R’ is varied
2. Current through R’ is nearly a constant as R’ is varied
3. CurrentI depends sensitively on R
4. I >= \(\frac{V}{r+R}\)always

Answer:

1. Potential drop across AB is nearly constant as R’ is varied

2. Current through R’ is nearly a constant as R’ is varied

Question 27. In a metre bridge point D is a neutral point.

1. The metre bridge can have no other neutral point for this set of resistances
2. When the jockey contacts a point on the metre wire left of D, current flows to B from the wire
3. When the jockey contacts a point on the metre wire to the right of D, current flows from B to the wire through the galvanometer
4. When R is increased, the neutral point shifts to the left

Answer:

1. The metre bridge can have no other neutral point for this set of resistances

3. When the jockey contacts a point on the metre wire to the right of D, current flows from B to the wire through the galvanometer

Question 28. The wires of lengths 3L and 4l are uniform. Then,

1. V_AB = 6V
2. V_BC = 3V
3. V_AC = 8V
4. V_CD = 4V

Answer:

1. V_AB = 6V

3. V_AC = 8V

4. V_CD = 4V

Question 29. If I = 5 A, then

1. V_AB = 30 V
2. V_CD = 8V
3. I₁ = 3 A
4. I₂ = 2A

Answer:

1. VAB = 30 V

3. I₁ = 3 A

4. I₂ = 2A

Question 30. If I₁ = 3 A and I₂ = 1 A, then

1. I₃ = 2 A
2. V_OA = 12V
3. V_OB = 2V
4. V_OC = 10V

Answer:

3. V_OB = 2V

4. V_OC = 10V

Kirchhoff’s laws class 12 MCQs 

Question 31. The null point of a meter bridge is at 40 cm when resistances X and Y are placed in the left and right gaps, respectively. The following observations show the resistances in the left and the right gaps and the null point respectively. Which of them is correct?

1. \(2 X, \frac{Y}{3}, 80 \mathrm{~cm}\)
2. 3X, 2Y, 50 cm
3. \(\frac{X}{2}, Y+\frac{X}{2}, 20 \mathrm{~cm}\)
4. \(X+\frac{Y}{2}, \frac{Y}{2}, 70 \mathrm{~cm}\)

Answer:

1. \(2 X, \frac{Y}{3}, 80 \mathrm{~cm}\)

2. 3X, 2Y, 50 cm

3. \(\frac{X}{2}, Y+\frac{X}{2}, 20 \mathrm{~cm}\)

4. \(X+\frac{Y}{2}, \frac{Y}{2}, 70 \mathrm{~cm}\)

Question 32. In the given circuit which options are correct?

1. The points A and B are at the same potential
2. The current through the battery is 5 A
3. The potential of A is 2.5 V higher than B
4. The potential of B is 2.5 V higher than A

Answer:

2. The current through the battery is 5 A

4. Potential of B is 2.5 V higher than A

Question 33. Considering the circuit mention the correct options.

1. Current passing through 2Ω is 2A
2. Current passing through 3Ω is 4 A
3. Current in the wire DE is zero
4. The potential of point A is 10 V

Answer:

1. Currentpassingthrough 2Ω is 2A

2. Current passing through 3Ω is 4 A

3. Potential of point A is 10 V

Question 34. In the network shown, choose the correct options.

1. I₁ = 3 A
2. I₂ = 2 A
3. The current through PQ is zero
4. The potential at S is less than at Q

Answer:

1. I₁ = 3 A

2. I₂ = 2 A

3. The current through PQ is zero

4. The potential at S is less than at Q

Kirchhoff’s laws class 12 MCQs 

Question 35. Between the Kirchhoff’s laws of electric circuits,

1. The First Law Signifies Conservation Of Charge
2. The First Law; Signifies Conservation Of Energy
3. The Second Law Signifies Conservation Of Charge
4. The second law signifies the conservation of energy

Answer:

1. The First Law Signifies Conservation Of Charge

4. The second law signifies conservation of energy

Question 35. A potentiometer is formed by applying a steady emf E0 across a uniform wire AB of length L. An electric cell has an emf E and an internal resistance r. The potential difference in an external circuit is V = E-Ir, where I is the current in that external circuit. The quantity E-V = Ir is called the internal drop of potential or the lost volt of the cell. If the cell is connected to the potentiometer wire through a galvanometer G, keeping the key K open, then for a certain point C, the galvanometer shows zero current; let CB = l. In this case, the lost volt of the cell would be zero as I = 0. Now, a shunt resistance R is connected in parallel to the cell and the key K is closed. Some current will then flow through the cell due to the shunt circuit and the lost volt will no longer be zero. Under this condition, the null point of the potentiometer circuit will shift from C to D,’ where DB = l'(l’ < l).

1. In the circuit, E₀ = 6 V and AB = 100 cm. When the key K is open, the null point is at C, with CB = 75 cm. The emf E of the cell is

1. 1.5V
2. 3.0V
3. 4.5V
4. 6.0V

Answer: 3. 4.5V

Practice MCQs on Circuit Analysis Using KCL and KVL

2. Next, a 2Ω resistance is used as the shunt R, and the key K is closed. If the null point shifts to a length of 60 cm (BD = 60 cm ), the value of r is

1. 0.5Ω
2. 1.0Ω
3. 1.5Ω
4. 2.0Ω

Answer: 1. 0.5Ω

3. If the 2Ω shunt resistance is replaced by 1Ω in the same circuit. Where will be the null point of the potentiometer when the key K is closed?

1. 42.5 cm
2. 50.0 cm
3. 57.5 cm
4. 65.0cm

Answer: 2. 50.0 cm

Question 36. The condition for balance of a Wheatstone bridge circuit, is \(\frac{P}{Q}=\frac{R}{S}\). Under this condition, no current passes through the galvanometer G. In that case, the galvanometer has no contribution to the equivalent resistance between points A and B and accordingly to the current I flowing through the circuit. So the resistance of G need not be taken into account in the calculations. Besides, the positions of the electric source E and of the galvanometer G are symmetric in the balanced condition. This means that the bridge retains its balance when they are exchanged to connect E between C and D and G between A and B.

1. In the circuit of, E = 3 V, P = 4Ω, Q = R = 6Ω and S = 9Ω. The current through P is

1. 0.20A
2. 0.30 A
3. 0.40A
4. 0.50 A

Answer: 2. 0.30 A

2. Now, if P and Q. are replaced by resistances of 6 n and 9Ω respectively, the current through P is

1. 0.20 A
2. 0.30 A
3. 0.40A
4. 0.50A

Answer: 1. 0.20 A

WBCHSE class 12 physics MCQs 

3. The arrangement is to be used as a Wheatstone bridge by connecting an electric source between O and C. The galvanometer is to be connected to the arm

1. OA
2. OB
3. AB
4. AC

Answer: 3. AB

4. The value of every resistance is R. The equivalent resistance between points A and B is

1. \(\frac{R}{2}\)
2. R
3. 2R
4. 4R

Answer: 2. R

WBCHSE class 12 physics MCQs 

Question 37. A wire of length 12 cm, resistance 12Ω, and of uniform area of cross-section is cut into twelve equal parts, which are connected to form a skeleton cube. A cell of emf 2V is connected across the two diagonally opposite comers of the cube. Using both Kirchhoff’s laws answer the following questions.

1. The effective resistance of the circuit is

1. \(\frac{4}{5} \Omega\)
2. \(\frac{5}{6} \Omega\)
3. \(\frac{6}{7} \Omega\)
4. \(\frac{7}{12} \Omega\)

Answer: 2. \(\frac{5}{6} \Omega\)

2. The current drawn from the battery is

1. 2.5 A
2. 2.4 A
3. 2.3 A
4. 3.4A

Answer: 2. 2.4 A

3. The maximum current flowing in an arm network is

1. 0.4 A
2. 0.8 A
3. 1.2 A
4. 2.4 A

Answer: 2. 0.8 A

4. The minimum potential difference across an arm of the network is

1. 0.4V
2. 0.8V
3. 1.2V
4. 2.4V

Answer: 1. 0.4V

WBCHSE class 12 physics MCQs 

Question 38. The current I in the circuit shown is

1. 1.33 A
2. zero
3. 2.00 A
4. 1.00 A

Answer: 1. 1.33 A

In the given circuit diagram, the clockwise circuit currents in the left and right circuits are fj and i2 respectively. According to Klrchhoff’s second law, for the left circuit,

2i₁ + 2(i₁ – i₂) = 4

or, 2i₁– i₂ = 2 …(1)

For the right circuit, 2(i₂– i₁) + 2i₂

= -4

or, – 2i₁ + 4i₂ = -4…(2)

Adding equations (1) and (2),

⇒ \(3 i_2=-2 \quad \text { or, } i_2=-\frac{2}{3} \mathrm{~A}\)

From equation (1),

⇒ \(2 i_1-\left(-\frac{2}{3}\right)=2 \text { or, } 2 i_1=\frac{4}{3} \quad \text { or, } i_1=\frac{2}{3} \mathrm{~A}\)

Hence, \(I=i_1-i_2=\frac{2}{3}-\left(-\frac{2}{3}\right)=\frac{4}{3} \mathrm{~A}=1.33 \mathrm{~A}\)

The option 1 is correct

Kirchhoff’s rules multiple choice questions 

Question 39. Two cells A and B emf 2 V and 1.5 V respectively, are connected as shown in through an external resistance 10Ω. The internal resistance of each cell is 5Ω. The potential difference E_A and E_B across the terminals of cells A and B respectively are

1. E_A = 2.0 V, E_B = 1.5 V
2. E_A = 2.125 V, E_B = 1.375 V
3. E_A = 1.875 V, E_B = 1.625 V
4. E_A = 1,875 V, E_B= 1.375 V

Answer: 3. E_A = 1.875 V, E_B = 1.625 V

Let the current I in the circuit flow in the anticlockwise direction.

According to Kirchhoff’s second law,

10I+5I+5I = 2-1.5

or, 20I = 0.5

or, I = 0.025 A

EA = E-Ir = 2- 0.025 x 5

= 1.875 V

EB = E-Ir

= 1.5- (-0.025) x 5

= 1.625 V

The option 3 is correct

Question 40 Consider the circuit where all the resistances are of magnitude 1 kΩ. If the current in the extreme right resistance X is 1mA, the potential difference between A and B is

1. 34V
2. 21V
3. 68V
4. 55V

Answer: 1. 34V

Here = 1 mA and X = 1Ω

The potential difference between 7 and 8

= i₂ – lkΩ

= V₇₈

= i1 x (l +l)

=1 mA x 2 kΩ

= 2 V

∴ \(i_2=\frac{2 \mathrm{~V}}{1 \mathrm{k} \Omega}=2 \mathrm{~mA}\)

and i₃ = i₁ + i₂

= 1 +2

= 3 mA

Similarly, the potential difference between 5 and 6,

V₅₆ = 5 V

∴ i₄ = 5 mA and i₅ = 8 mA

In the same way, the potential difference between 3 and 4,

V₃₄ = 13 V

∴ i₆ = 13 mA and i₇ = 21 mA

The potential difference between 1 and 2,

V₁₂ = 13 + 21 = 34 V

∴ V₁₂ = V_AB

hence, V_AB = 34 V

The option 1 is correct

Kirchhoff’s rules multiple choice questions 

Question 41. Consider the circuit given here. The potential difference V_BC between points B and C is

1. IV
2. 0.5V
3. 0V
4. -1V

Answer: 2. 0.5V

⇒ \(i=\frac{V}{R}=\frac{3}{6 \times 10^3}=0.5 \times 10^{-3} \mathrm{~A}\)

⇒ \(V_{A D}=i R=\left(0.5 \times 10^{-3}\right) \times(1+2) \times 10^3=1.5 \mathrm{~V}\)

Equivalent capacitance,

⇒ \(C=\frac{1}{1+\frac{1}{2}}=\frac{2}{3} \mu \mathrm{F}\)

∴ The charge stored in the combination of capacitors,

⇒ \(Q=C V=\frac{2 \times 1.5}{3}=1 \mu \mathrm{C}\)

Applying Kirchhoff’s second law to the loop BDCB,

⇒ \(V_B-V_C=i R-\frac{Q}{C}=\left(0.5 \times 10^{-3} \times 2 \times 10^3\right)-\frac{1}{2}=0.5 \mathrm{~V}\)

The option 2 is correct.

Important Definitions in Electrical Measurements

Question 42. A non-zero current passes through the galvanometer G shown in the circuit when the key K is closed and its value does not change when the key is opened. Then which of the following statement(s) is/are true?

1. The galvanometer resistance is infinite.
2. The current through the galvanometer is 40 mA.
3. After the key is closed, the current through the 200Ω resistor is the same as the current through the 300Ω resistor.
4. The galvanometer resistance is 150Ω

Answer: 2, 3 and 4

Since the same current flows through the galvanometer for both cases when switch K is closed or open, so the Wheatstone bridge.

⇒ \(\frac{200}{300}=\frac{100}{G}\) [G = resistance of galvanometer]

or, \(G=\frac{150 \times 300}{200}=150 \Omega\)

When the switch is open, the current through the galvanometer

= \(\frac{10}{100+150} A=40 \mathrm{~mA}\)

Again, when the switch K is closed, the equivalent circuit is shown in the figure below.

∴ Equivalent resistance

⇒ \(\frac{200 \times 100}{300}+\frac{300 \times 150}{450}=166 \frac{2}{3} \Omega\)

⇒ \(\text { Current, } I=\frac{10}{166 \frac{2}{3}} \mathrm{~A}\)

∴ Current through 200Ω resistor

⇒ \(\frac{10}{166 \frac{2}{3}} \times \frac{100}{300}=\frac{10}{166 \frac{2}{3} \times 3} \mathrm{~A}\)

and current through 300Ω resistor

⇒ \(\frac{10}{166 \frac{2}{3}} \times \frac{150}{450}=\frac{10}{166 \frac{2}{3} \times 3} \mathrm{~A}\)

The options 2, 3, and 4 are correct.

Kirchhoff’s rules multiple choice questions 

Question 43. In the circuit shown the current in the lΩ resistor is

1. 1.3 A, from P to Q
2. 0A
3. 0.13 A, from Q to P
4. 0.13 A, from P to Q

Answer: 3. 0.13 A, from Q to P

Let the clockwise circuit currents in the left and right circuits be I1 and I2, respectively.

For the left circuit, 3I₁ +l(I₁– 12) = -6

or, 4I₁ – I2 = -6….(1)

For the right circuit, 3I₂ + 2I₂ + I(I₂-I₁) = -9

or, I₁– 6I₂ = 9

Solving equations (1) and (2), we get

I₁ = -1.96A A and I₂ = -1.83 A

∴ I2-I₁ = -1.83 -(-1.96)

= 0.13 A

= current in the lΩ resistance, from Q to P

The option 3 is correct.

Question 44. In the given circuit the current in each resistance is

1. 1A
2. 0.25 A
3. 0.5 A
4. Zero

Answer: 4. Zero

Each loop contains two cells of equal and opposite emf. So the net emf in each loop is zero and hence the current in each resistance is zero.

The option 4 is correct.

Examples of Kirchhoff’s Laws Applications

Question 45. Which of the following statements is false?

1. Wheat stone bridge is the most sensitive when all four resistances are of the same order of magnitude.
2. In a balanced Wheatstone bridge, if the cell and the galvanometer are exchanged, the null point is disturbed.
3. A rheostat can be used as a potential divider.
4. Khchhoff’s second law represents energy conservation.

Answer: 2. In a balanced Wheatstone bridge if the cell and the galvanometer are exchanged, the nullpoint is disturbed.

If the positions of the cell and the galvanometer are exchanged in a Wheatstone bridge, the null condition remains unchanged.

The option 2 is correct.

Question 46. In a potentiometer experiment, it is found that no current passes through the galvanometer when the terminals of the cell are connected across 52 cm of the potentiometer wire. If the cell is shunted by a resistance of 5Ω, a balance is found when the cell is connected across 40 cm of the wire. Find the internal resistance of the cell

1. 2Ω
2. 2.5Ω
3. 1Ω
4. 1.5Ω

Answer: 4. 1.5Ω

When a potentiometer is used for the determination of the internal resistance of a cell,

⇒ \(r=\left(\frac{l_1}{l_2}-1\right) R\)

[r is the internal resistance of the cell, R is the known
resistance l1 and l2 are the positions of the null point in the
two cases]

∴ \(r=\left(\frac{52}{40}-1\right) \times 5=1.5 \Omega\)

The option 4 is correct.

Class 12 physics circuit analysis MCQs 

Question 47. Two batteries with emf 12 V and 13 V are connected in parallel across a load resistor of 10 Ω. The internal resistances of the two batteries are 1Ω and 2Ω respectively. The voltage across the load lies between

1. 11.4 V and 11.5 V
2. 11.7 V and 11.8 V
3. 11.6 V and 11.7 V
4. 11.5 V and 11.6 V

Answer: 4. 11.5 V and 11.6 V

Applying Kirchhoff’s second law,

12 – x × 1 – 10(x+ y) = 0 [x and y are currents]

or, 12 = 11x + 10y ….(1)

Similarly, 13 = 10x + 12y …(2)

From equation (1) and (2) we get,

⇒ \(x=\frac{7}{16} \mathrm{~A}, y=\frac{23}{32}\)

∴ \(V=10\left(\frac{7}{16}+\frac{23}{32}\right)\)

= 11.56V

The option 4 is correct

Question 48. On interchanging the resistances, the balance point of a metre bridge shifts to the left by 10 cm. The resistance of their series combination is 1 kΩ. How much was the resistance on the left slot before interchanging the resistances?

1. 550Ω
2. 910Ω
3. 990Ω
4. 505Ω

Answer: 1. 550Ω

The balance condition in the first case,

⇒ \(\frac{R_1}{l}=\frac{R_2}{100-l}\)

or, \(\frac{R_1}{R_2}=\frac{l}{100-l}\)

When the resistances Rt and R2 are interchanged in the second case,

⇒ \(\frac{R_2}{R_1}=\frac{l-10}{100-(l-10)}\)

or, \(\frac{R_1}{R_2}=\frac{110-l}{l-10}\)…(2)

From equation (1) and (2) we get,

⇒ \(\frac{l}{100-l}=\frac{110-l}{l-10}\)

or, l(l — 10) = (100-l)(110-l)

or, l = 55 cm

From equation (1) we get,

⇒ \(R_1=\frac{55}{45} R_2\)

Given, R₁ + R₂ = 1000

∴ \(R_1=\frac{55}{45}\left(1000-R_1\right) \quad \text { or, } R_1=550 \Omega\)

The option 1 is correct.

Real-Life Scenarios in Electrical Measurement

Question 49. A circuit has been set up to find the internal resistance of a given cell. The main battery used across the potentiometer wire has an emf of 2.0 V and negligible internal resistance. The potentiometer wire itself is 4m long. When the resistance R, connected across the given cell, has values of

1. Infinity,
2. 9.5Ω the balancing lengths on the potentiometer wire are found to be 3m and 2.85m respectively.

The value internal resistance of the cell is

1. 0.25Ω
2. 0.95Ω
3. 0.Ω
4. 0.75Ω

Answer: 3. 0.5Ω

The internal resistance of the cell,

⇒ \(r=\left(\frac{l_1}{l_2}-1\right) R=9.5\left(\frac{3}{2.85}-1\right)=0.5 \Omega\)

The option is correct

Question 50. In an ammeter, 0.2% of the main current passes through the galvanometer. If the resistance of the galvanometer is G, the resistance of the ammeter will be

1. \(\frac{1}{499}\)G
2. \(\frac{499}{500}\)G
3. \(\frac{1}{500}\)G
4. \(\frac{500}{499}\)G

Answer: 3. \(\frac{1}{500}\)G

⇒ \(S=\frac{I_G}{I_S} \cdot G=\frac{I_G}{I-I_G} \cdot G=\frac{\frac{0.2}{100}}{1-\frac{0.2}{100}} G\)

⇒ \(\frac{0.2}{100} \times \frac{100}{99.8} G=\frac{1}{499} G\)

∴ Resistance of the ammeter

⇒ \(=\frac{S G}{S+G}=\frac{\frac{1}{499} \cdot G \cdot G}{\frac{1}{499} G+G}=\frac{\frac{1}{499} G}{\frac{500}{499}}=\frac{1}{500} G\)

The option 3 is correct.

Question 51. The resistance in the two arms of the meter bridge is 5Ω and RΩ respectively. When the resistance R is shunted with an equal resistance, the new balance at 1.6Ω. The resistance R is

1. 10Ω
2. 15Ω
3. 20Ω
4. 25Ω

Answer: 2. 15Ω

In first case, \(\frac{5}{l_1}=\frac{R}{100-l_1}\)

In second case, \(\frac{5}{1.6 l_1}=\frac{\frac{R}{2}}{100-1.6 l_1}\)

∴ \(\frac{R \times 1.6}{2\left(100-1.6 l_1\right)}=\frac{R}{\left(100-l_1\right)} \text { or, } l_1=25\)

So, \(R=\frac{5}{l_1} \times\left(100-l_1\right)=\frac{5}{25} \times 75=15 \Omega\)

Option 2 Is correct.

Question 52. A potentiometer wire has a length of 4 m and a resistance of 8Ω. The resistance that must be connected in series with the wire and an accumulator of emf 2 V, so as to get a potential gradient 1 mV per cm on the wire is

1. 32Ω
2. 40Ω
3. 44Ω
4. 48Ω

Answer: 1. 32Ω

Resistance of the potentiometer wire of length 1 cm

⇒ \(\frac{8 \Omega}{400 \mathrm{~cm}}=\frac{1}{50} \Omega / \mathrm{cm}\)

If the potential difference at the two ends of the 1 cm portion is 1 mV or 10-3V

then the current through the wire \(I=\frac{10^{-3}}{\frac{1}{50}}=\frac{1}{20} \mathrm{~A}\)

For the entire circuit, \(I=\frac{E}{R_P+R}\)

or, \(R=\frac{E}{I}-R_P=\frac{2}{\frac{1}{20}}-8=32 \Omega\)

The option 1 is correct

Class 12 physics circuit analysis MCQs 

Question 53. A potentiometer wire is 100 cm long and a constant potential difference is maintained across it Two cells are connected in series first to support one another and then in opposite directions. The balance points are obtained at 50 cm and 10 cm from the positive end of the wire in the two cases. The ratio of emf’s is

1. 5:4
2. 3:4
3. 3:2
4. 5:1

Answer: 3. 3:2

The emf of the two cells are E₁ and E₂ then,

E₁ + E₂ = 50k

and E₁ – E₂ = 10k [where k = constant]

∴ \(\frac{E_1+E_2}{E_1-E_2}=\frac{5}{1}\)

or, \(\frac{E_1}{E_2}=\frac{5+1}{5-1}=\frac{3}{2}\)

The option 3 is correct

Class 12 physics circuit analysis MCQs 

Unit 2 Current Electricity Chapter 2 Kirchhoff’s Laws And Electrical Measurement Synopsis

• Kirchhoff’s current law: The algebraic sum of all currents through the conductors meeting at any point in a circuit is zero.
• Kirchhoff’s voltage law: In a closed loop of an electrical circuit, the algebraic sum of the products of each resistance and the associated current through them is equal to the algebraic sum of the electromotive forces present in that loop.
• Uses of potentiometer: A potentiometer is used
  • As a variable resistor,
  • As a source of variable emf,
  • For the determination of the emf of a cell
  • To measure the internal resistance of a cell
  • To measure the potential difference between any two points in an electrical circuit.
• Conditions for sensitivity Wheatstone Bridge:
  • The galvanometer should be sensitive so that even for a minute current flowing through the galvanometer, its indicator shows deflection and
  • The resistances of the four arms of the bridge and that of the galvanometer should approximately be equal in magnitude.
• For the measurement of very high resistance (< 1Ω) and also of very high resistance (> 100Ω), the Wheatstone bridge cannot be used.
• In the null condition of Wheatstone bridge,

• \(\frac{P}{Q}\) = \(\frac{R}{S}\) and in this condition, I_G = 0
• The formula for determination of unknown resistance with the help of a metre bridge:

⇒ \(S=R\left(\frac{100-l}{l}\right)\)

• When the unknown resistance S is connected in the right gap of the metre bridge and l is the position of the null point on the metre wire.
• In the circuit given below if the potential of the points A, B and C are V₁, V₂ and V₃ then the potential of point O will be

⇒ \(V_O=\left(\frac{V_1}{R_1}+\frac{V_2}{R_2}+\frac{V_3}{R_3}\right)\left(\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\right)^{-1}\)

• If R be the value of each resistance and V is the potential difference between the points A and B in the circuit given below then,
  • Equivalent resistance between the points A and B is R_AB = R
  • Current through AF (or EB), I = \(\frac{V}{R}\)
  • Current through AFCEB (or AFDEB), I = \(\frac{V}{2R}\)

• In the balanced condition, the equivalent resistance between the points A and C of the Wheatstone bridge,

⇒ \(R_{A C}=\frac{(P+Q)(R+S)}{P+Q+R+S}\)