WBCHSE Class 12 Physics Kirchhoff’s Laws And Electrical Measurement Question and Answers

Kirchhoff’s Laws And Electrical Measurement Long Questions And Answers

Question 1. What is the potential at point O of the circuit?

Answer:

If V is the potential of the point 0, then flowing towards 0 through each 5Ω resistance = \(\frac{2-V}{5} \mathrm{~A}\).

Applying,

Kirchhoff’s current law at the point O we have,

⇒ \(\frac{2-V}{5}+\frac{2-V}{5}+\frac{2-V}{5}=0\)

or, \(\frac{6-3 V}{5}=0\)

or, V = 2V

So, no current will flow through any resistance.

Question 2. If the points B and C in the circuit are earthed, what is the current flowing through each 5Ω resistance?

Answer:

The two points B and C are earthed. So potentials of these two points are zero. If V is the potential of the point 0, then

Current flowing along \(A O=\frac{2-V}{5} \mathrm{~A}\)

Current flowing along \(B O=\frac{0-V}{5} \mathrm{~A}=-\frac{V}{5} \mathrm{~A}\)

Current flowing along \(C O=\frac{0-V}{5} \mathrm{~A}=-\frac{V}{5} \mathrm{~A}\)

Applying Kirchhoff’s first law at point 0 we have,

⇒ \(\frac{2-V}{5}-\frac{V}{5}-\frac{V}{5}=0 \quad \text { or, } \frac{2-3 V}{5}=0\)

or, V = \(\frac{2}{3}[latex] V

∴ Current flowing along [latex]A O=\frac{2-\frac{2}{3}}{5}=\frac{4}{15} \mathrm{~A}\),

Current flowing along \(O B=\frac{\frac{2}{3}}{5}=\frac{2}{15} \wedge\)

Current flowing along \(O C=\frac{\frac{2}{3}}{5}=\frac{2}{15} \mathrm{~A}\)

WBBSE Class 12 Kirchhoff’s Laws Q&A

Question 3. On what principle does a potentiometer work?
Answer:

The main principle of a potentiometer is as follows: The resistance of any two portions of the potentiometer wire having equal lengths is equal. So, while current flows through the potentiometer wire, everywhere the potential drop for equal lengths is equal.

Question 4. The readings of the two ammeters A₁ and A₂ in the circuit are 1.5 A and 1.0 A respectively. What is the current through the resistance R?

Answer:

Currents of the two loops are taken as i₁ and i₂.

So, i₁ = 1.5 A and i₁– i₂ = 1 A

∴ i₂ = i₁-l

= 1.5-1

= 0.5 A

= current through the resistance R

Question 5. The reading of the ammeter in the circuit is zero. What is the reading of the voltmeter?

Answer:

Since the reading is zero, obviously no current flows through the battery of emf e₂. So for this battery, lost volt = 0. Therefore, the voltmeter indicates the emf of this battery i.e., reading of the voltmeter = e₂.

Key Concepts in Kirchhoff’s Laws Questions

Question 6. Why is a potentiometer preferred to a voltmeter for the measurement of emf of a cell? Explain.
Answer:

Let the emf of tire cell =E and its internal resistance =r. When the cell provides current 7 in the closed circuit, the potential difference across the cell, V = E-Ir. When a voltmeter is connected across the cell, we get a reading for the potential difference V which is less than the actual emf, E of the cell. On the other hand, the circuit is set up in such a way for the potentiometer system so that no current flows through the cell, i.e., 7=0. Hence, V = E. Therefore, a potentiometer is preferred to a voltmeter for the measurement of emf of a cell.

Question 7. Can we apply Kirchhoff’s laws in the circuit having non-ohmic conductors?
Answer:

Both the laws of Kirchhoff are applicable to circuits having non-ohmic conductors. But in that case, for the determination of current or potential difference when we apply the relation V = IR, we should remember that the value of R is not constant. In each case, we shall have to know the correct relation between V and 7 and then with the help of Kirchhoff’s laws, the analysis of the circuit is possible.

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Question 8. What type of cell should be used in Wheatstone Bridge?
Answer:

The null condition of the Wheatstone bridge does not depend on the emf of the cell. In spite of that, the cell which can send high current should not be used. In that case, the resistance of each arm of the bridge may increase considerably due to the generation of heat by the Joule effect.

So, storage cells are not used in a Wheatstone bridge circuit. It is better to use a Leclanche cell because its emf is not very high and the current sends also remains sufficiently low

Question 9. Why is it not possible to measure the emf of a cell correctly by a voltmeter? Under what conditions correct measurement is possible?
Answer:

Suppose, the emf of a cell -E and its internal resistance =r. If the cell sends Current I in a closed circuit, then the terminal potential differential, of the cell is V = E-Ir. If a voltmeter is connected to the terminals of the cell, it reads the terminal potential difference, which is less than the emf, of the cell. The condition for obtaining the correct measurement of emf is

V = E i.e., lost volt Ir = 0. This condition may be fulfilled under any of the following two circumstances,

1. The internal resistance of the cell r = 0
2. The current through the cell, I = 0.

When the second condition is fulfilled, correct measurements of the emf of a cell is possible. Practically, it is not possible to fulfil the first condition.

Question 10. A cell of emf V volts and negligible internal resistance is connected across the potentiometer whose sliding contact is placed exactly in the middle. A voltmeter is connected between the sliding contact and one fixed end of the potentiometer. If It Is assumed that the resistance of the voltmeter is not very high compared with the resistance of the potentiometer, what voltage will the voltmeter show higher than or less than \(\frac{V}{2}\)?
Answer:

B is the midpoint of the potentiometer AC

So long as the voltmeter D is not connected,

V_AB = V_BC = \(\frac{V}{2}\).

But as soon as the voltmeter is connected between points B and C the resistance of the portion BC and the resistance of the voltmeter D form a parallel combination.

So, the equivalent resistance between the points B and C becomes less than the resistance of BC, i.e., in this condition the resistance of AB is greater than the resistance of BC.

So V_AB > V_BC i.e., the potential difference between A and B will be greater than \(\frac{V}{2}\) and that between B and C will be less than. So, the voltmeter will record a reading less than \(\frac{V}{2}\).

Short Answer Questions on Kirchhoff’s Current Law

Question 11. Will the position of the null point change if the galvanometer is replaced by another one of a different resistance in a Wheatstone bridge?
Answer:

The condition of balance i.e., \(\frac{P}{Q}\) = \(\frac{R}{S}\) does not depend on the resistance of the galvanometer. So, if a galvanometer of a different resistance is used, the position of the null point does not change. Of course, the sensitivity of the bridge depends on the resistance of the galvanometer.

Question 12. How will the position of the null point of a Wheatstone bridge change if we interchange the positions of the battery and the galvanometer in the circuit?
Answer:

Suppose, the battery is connected between points A and B of the Wheatstone bridge and the galvanometer is connected between points C and D, Then the null condition of the bridge is

⇒ \(\frac{P}{Q}\) = \(\frac{R}{S}\)

Next die battery is connected between points C and D and the galvanometer is connected between points A and B. In this case die resistances of the first, second, third and fourth arms of the bridge are. R, P, S and Q respectively. So the null condition of the bridge now is,

⇒ \(\frac{R}{P}\) = \(\frac{S}{Q}\)

or, \(\frac{P}{Q}\) = \(\frac{R}{S}\)

Obviously, in both cases, the null condition is exactly the same, i.e., due to the interchange of battery and the galvanometer the null condition of the bridge does not change.

Question 13. In the circuit given in the, P ≠ R. Irrespective of whether the switch S is open or closed, the reading of the galvanometer remains unaltered. Select the correct answer from the following statements

1. I_R = I_Q
2. I_P = I_Q
3. I_Q = I_G
4. I_Q = I_R

Answer:

The closing of switch S does not affect the current in the galvanometer, i.e., I_S = 0. So, the current passing through G will be the same as that passing through the resistance R.

Hence, I_R = I_G, so the statement 1 is correct

Question 14. Can we compare the two resistances 1Ω and 100Ω accurately with the help of a metre bridge?
Answer:

Metre Bridge works on the principle of a Wheatstone bridge. We know that the bridge becomes very sensitive when the resistances of the four arms are nearly equal. Then the resistances can be compared accurately. Obviously, 1Ω and 100Ω cannot be’ compared accurately with the help of a metre bridge.

Question 15. Sometimes the balance point in the potentiometer may not be obtained on the wire of the potentiometer. Under what conditions does it happen?
Answer:

Let a cell of emf B be connected across the entire length L of a potentiometer wire. Now, if the balance point is obtained at a length l during the measurement of an unknown voltage V, then \(\frac{E}{V}=\frac{L}{l}\)

The balance point is not on the potentiometer wire – this statement means that l>L.

In that case, V> E

Common Questions on Kirchhoff’s Voltage Law

Question 16. Three resistances R₁, R₂ and R₃ are connected in parallel. This combination is then connected to a cell of negligible internal resistance. Applying Kirchhoff’s law proves that the equivalent resistance of the whole combination is given by,

⇒ \(R=\frac{R_1 R_2 R_3}{R_1 R_2+R_2 R_3+R_1 R_3}\)

Answer:

Applying Kirchhoff’s second law in the closed loop AR₁BEA,

i₁R₁ – E = 0 [where the cell is E and internal resistance is zero]

∴ \(i_1=\frac{E}{R_1}\)…(1)

Again applying Kirchhoff’s second law in AR₂BEA,

i₂R₂ – E = 0

∴ \(i_2=\frac{E}{R_2}\)…(2)

and similarly,

From equations (l), (2) and (3) we get,

⇒ \(i_1+i_2+i_3=E\left[\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\right]=i[i=\text { total current }]\)

If R be the equivalent resistance then,

⇒ \(R=\frac{E}{i}=\frac{1}{\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}}=\frac{R_1 R_2 R_3}{R_1 R_2+R_2 R_3+R_1 R_3}\)

Question 17. The variation of potential difference V with length f In the case of two potentiometers X and Y. Which one of the two will you prefer for comparing emfs of the two cells and why?

Answer:

A potentiometer is said to be sensitive if the potential drop per unit length, i.e., potential gradient \(\frac{dV}{dl}\) is small. From the given graph,

the slope of Y < slope of X

∴ \(\left(\frac{d V}{d l}\right)_Y<\left(\frac{d V}{d l}\right)_X\)

∴ Potentiometer Y will be preferred for comparing emfs of the two cells.

Conceptual Questions on Circuit Theory Using KCL and KVL

Question 18. Find the potential difference between the left and right plates of each capacitor in the circuit. (Assume, E₂ > E₁ )

Answer:

The right plate of C₁ has charge +q and the left plate of C₁ has charge -q.

Similarly, the left plate of C₂ has charge + q and the right plate of C₂ has charge -q.

Applying Klrchhoff’s second Inw to the closed loop we get

⇒ \(\frac{q}{C_1}+B_1+\frac{q}{C_2}-B_2=0\)

or, \(q=\frac{\left(B_2-B_1\right) C_1 C_2}{C_1+C_2}\)

Hence, the potential differences across the left and right plates of C₁

⇒ \(v_1=\frac{q}{C_1}=\frac{\left(E_2-E_1\right) C_2}{\left(C_1+C_2\right)}\)

Similarly, the potential difference across left and right plate of C₂

⇒ \(V_2=\frac{q}{C_2}=\frac{\left(E_2-E_1\right) C_1}{\left(C_1+C_2\right)}\)

Question 19. In the given circuit, determine the condition for which V_A-V_B = 0.

Answer:

The capacitors In the series have die same charge. So applying Klrchhoff’s second law to the loop containing C₁, C₂ and E we get,

⇒ \(\frac{q}{C_1}+\frac{q}{C_2}-E=0\)

or, \(q=E\left[\frac{C_1 C_2}{C_1+C_2}\right]\)

Similarly, applying Klrchhoff’s second law to the loop containing C₃, C₄ and E we get,

⇒ \(\frac{q^{\prime}}{C_3}+\frac{q^{\prime}}{C_4}-E=0\)

or, \(q^{\prime}=E\left[\frac{C_3 C_4}{C_3+C_4}\right]\)

Now, \(V_A-V_B=\frac{q}{C_2}-\frac{q^{\prime}}{C_4}=E\left[\frac{C_1}{C_1+C_2}-\frac{C_3}{C_3+C_4}\right]\)

= \(E\left[\frac{C_1 C_4-C_3 C_2}{\left(C_1+C_2\right)\left(C_3+C_4\right)}\right]\)

For, \(V_A-V_B=0, C_1 C_1-C_2 C_3=0\)

or, \(\frac{C_1}{C_2}=\frac{C_3}{C_4}\)