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

Kirchhoff’s Laws And Electrical Measurement Short Question And Answers

Question 1. Determine the current in each branch of the network

Class 12 Physics Unit 2 Current Electricity Chapter 2 Kirchhoff’s Laws And Electrical Measurement the current in each branch of the network

Answer:

Applying Kirchhoff’s second law to the network ABDA, we get,

⇒ \(10 I_1+5 I_2-5\left(I-I_1\right)=0 \quad \text { or, } 3 I_1+I_2-I=0\)….(1)

For the loop BCDB,

⇒ \(5\left(I_1-I_2\right)-10\left(I-I_1+I_2\right)-5 I_2=0\)

or, 3I1-4I2-27 =0….(2)

For the loop ABCGHA,

⇒ \(10 I+10 I_1+5\left(I_1-I_2\right)=0\)

or, 2I+3I1-I2 = 2….(3)

Solving the equations we get,

⇒ \(I=\frac{10}{17} \mathrm{~A}, I_1=\frac{4}{17} \mathrm{~A}, I_2=-\frac{2}{17} \mathrm{~A}\)

∴ Current through \(A B=\frac{4}{17} A\)

Current through \(B C=\frac{4}{17}+\frac{2}{17}=\frac{6}{17} \mathrm{~A}\)

Current through \(D C=\left(I-I_1+I_2\right)\)

⇒ \(=\frac{10}{17}-\frac{4}{17}-\frac{2}{17}=\frac{4}{17} \mathrm{~A}\)

Current through \(B D=\frac{2}{17} A\)

Current through \(A D=\left(I-I_1\right)=\frac{10-4}{17}=\frac{6}{17} \mathrm{~A}\)

Question 2. A potentiometer with a cell of 2.0 V internal resistance 0.40Ω maintains a potential drop across the resistor wire AB. A standard cell that maintains a constant emf of 1.02 V (for very moderate currents up to a few) gives a balance point at 67.3 cm length of the wire. To ensure very low currents drawn from the standard cell, a very high resistance of 600 kΩ is put in series with it, which is shorted close to the balance point. The standard cell is then replaced by a cell of unknown emf e and the balance point found similarly, turns out to be at 82.3 cm length of the wire.

Class 12 Physics Unit 2 Current Electricity Chapter 2 Kirchhoff’s Laws And Electrical Measurement potentiometer with a cell

1. What is the value of e?

2. What purpose does the high resistance of 600 kΩ serve?

3. Is the balance point affected by this high resistance?

4. Is the balance point affected by the internal resistance of the driver cell?

5. Would the method work in the above situation if the driver cell of the potentiometer had an emf of 1.0 V instead of 2.0 V?

6. Would the circuit work well for determining an extremely small emf, say of the order of a few mV? If not, how will you modify the circuit?

Answer:

1. Using \(E \propto l, \frac{E_2}{E_1}=\frac{l_2}{l_1}\)

or, \(E_2=\frac{l_2}{l_1} \cdot E_1=\frac{82.3}{67.3} \times 1.02=1.25 \mathrm{~V}\)

2. The high resistor allows a very small current to flow
through the galvanometer when the circuit is not
balanced.

3. The balance point is not affected by the high resistance.

4. The internal resistance of the driver cell has no influence on the balance point.

5. If the emf of the driver cell is IV instead of 2V, no balance point will be obtained. Hence the arrangement will not work.

6. For determining extremely small emf, the balance point will be almost on end and the measurement will not be accurate.

For effective measurement tlÿVemf of the driver cell should be very small.

Question 3. Two resistances are compared by a potentiometer. The balance point with a standard resistor R = 10.0Ω is found to be 58.3 cm, while that with the unknown resistance X is 68.5 cm.

Class 12 Physics Unit 2 Current Electricity Chapter 2 Kirchhoff’s Laws And Electrical Measurement Two resistances are compared by a potentiometer

  1. What is the value of X?
  2. What would you do if no balance point is obtained using the given cell of emf E?

Answer:

1. \(\frac{E_2}{E_1}=\frac{X}{R}\)

or, \(X=R \cdot \frac{E_2}{E_1}=R \cdot \frac{l_2}{l_1}\)

or, \(X=10 \times \frac{68.5}{58.3}=11.75 \Omega\)

2. In order to obtain a balance point, either a cell of emf (E’) less than E has to be used or a suitable resistor has to be put in series with R and X so as to reduce the potential drop across AB.

Question 4. 2.0 V potentiometer used for the determination of internal resistance of a 1.5 V cell. The balance point of the cell in an open circuit is 76.3 cm. When a resistor of 9.5Ω is used in the external circuit of the cell, the balance point shifts to the 64.8 cm length of the potentiometer wire. Determine the internal resistance of the cell

Class 12 Physics Unit 2 Current Electricity Chapter 2 Kirchhoff’s Laws And Electrical Measurement the internal resistance of the cell

Answer:

⇒ \(\left(l_1-l_2\right) R=l_2 r\)

∴ \(r=\left(\frac{l_1}{l_2}-1\right) R\) [Here, l1 = 76.3 cm, l2 = 64.8 cin, R = 9.5Ω]

⇒ \(\left(\frac{76.3}{64.8}-1\right) \times 9.5=1.68 \Omega\)

Question 5. Establish the balanced condition of Wheatstone’s bridge by applying Klrchhoff’s laws,
Answer:

In balanced condition, IG = 0.

Therefore, by using Klrchhoff’s second law,

for the loop ABDA, l1P-l2R = 0

or, \(\frac{I_1}{I_2}=\frac{R}{P}\)

For the loop BCDB, \(I_1 Q-I_2 S=0 \quad \text { or, } \frac{I_1}{I_2}=\frac{S}{Q}\)

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

This is the balanced condition of Wheatstone’s bridge.

Question 6. How can the sensitivity of a potentiometer be increased?
Answer:

The sensitivity of a potentiometer is increased if the null point is formed near the midpoint of the potentiometer wire.

Question 7. A potentiometer has 10 wires each 1 metre in length and the total resistance is 20Ω. Find the resistance to be connected to the driving battery of emf 2 volts to produce a potential drop of 1μV per millimeter. (Graph sheet is not required).
Answer:

The total length of the 10 wires = 1 x 10

= 10 m

Potential drop = 1μV/mm = \(\frac{10^{-6}}{10^{-3}} \mathrm{~V} / \mathrm{m}\)

= 10-3V/m

So, the potential drop across the whole wire

= 10-3 x 10

= 10-2 V

The resistance of the whole wire = 20Ω

∴ Currently the potentiometer wire

⇒ \(\frac{10^{-2}}{20}=0.5 \times 10^{-3} \mathrm{~A}\)

∴ Net resistance of the circuit = \(\frac{2}{0.5 \times 10^{-3}}=4000 \Omega\)

∴ The resistance connected to the driving battery

= 4000-20

= 3980Ω

Question 8. Determine the value of I in the circuit

Class 12 Physics Unit 2 Current Electricity Chapter 2 Kirchhoff’s Laws And Electrical Measurement the value of I in tbe circuit

Answer:

Let VA = VB = VC = 0 and VF = V

Hence, VD = VE = 2; VG = V+2

Class 12 Physics Unit 2 Current Electricity Chapter 2 Kirchhoff’s Laws And Electrical Measurement the value of I in tbe circuit.

Applying Kirchhoff’s first law at the point F,

⇒ \(\frac{V-2}{2}+\frac{V-2}{2}+I=0 \quad \text { or, } I=2-V\)…(1)

Again, along with GB,

⇒ \(I=\frac{(V+2)-0}{2} \text { or, } 2 I=V+2\)….(2)

From equations (1) and (2) we get

⇒ \(3 I=4 \quad \text { or, } I=\frac{4}{3} \mathrm{~A}\)

Question 9. Calculate the value of the resistance R in the circuit. So that the current in the circuit is 0.2 A. What would be the potential difference between points B and E?

Class 12 Physics Unit 2 Current Electricity Chapter 2 Kirchhoff’s Laws And Electrical Measurement potential difference between points B and E

Answer:

Resistanceinbranch BCD = 10 + 5 = 15Ω

Let r = equivalent resistance between B and E.

Then, \(\frac{1}{r}=\frac{1}{30}+\frac{1}{10}+\frac{1}{15}\)

= \(\frac{1+3+2}{30}\)

= \(\frac{6}{30}\)

= \(\frac{1}{5}\)

i.e., r = 5Ω

The potential difference between points B and E

=Ir = 0.2 x 5

= 1V

Effective emf of the circuit = 8-3

= 5 V

∴ \(0.2=\frac{5}{15+R+5}=\frac{5}{20+R}\)

or 4 + 0.2R = 5

or, \(R=\frac{5-4}{0.2}=5 \Omega\)

Question 10. Two identical cells, each of emf E, having negligible internal resistance, are connected in parallel with each other across an external resistance R. What is the current through this resistance?
Answer:

The cells are in parallel; so the voltage applied on R = E.

The current through R is I = \(\frac{E}{R}\)

Question 11. Two circuits each having a galvanometer and a battery of 3 V. When the galvanometers in each arrangement do not show any deflection, obtain the ratio R1/R2

Class 12 Physics Unit 2 Current Electricity Chapter 2 Kirchhoff’s Laws And Electrical Measurement galvanometers

Answer:

In the first arrangement

⇒ \(\frac{4}{R_1}=\frac{6}{9} \quad \text { or, } R_1=\frac{4 \times 9}{6}=6 \Omega\)

In the second arrangement, \(\frac{12}{8}=\frac{6}{R_2} \quad \text { or, } \frac{6 \times 8}{12}=4 \Omega\)

So, \(\frac{R_1}{R_2}=\frac{6}{4}=\frac{3}{2}\)

Question 12.

  1. Why are the connections between the resistors in a meter bridge made of thick copper strips?
  2. Why is it generally preferred to obtain the balance point
    in the middle of the meter bridge wire?
  3. Which material is used for the meter bridge wire and why?

Answer:

1. The connections between the resistors in a meter bridge are made of thick copper strips because of their negligible resistance.

2. It is generally preferred to obtain the balance point in the middle of the meter bridge wire because the meter bridge is most sensitive when all four resistances are of the same order.

3. Alloy, manganin, or constant are used for making meter bridge wire due to the low-temperature coefficient of resistance and high resistivity

Question 13. A resistance of RΩ draws current from a potentiometer. The potentiometer has a total resistance of R0Ω. A voltage V is supplied to the potentiometer. Derive an expression for the voltage across R when sliding contact is in the middle of the potentiometer.

Class 12 Physics Unit 2 Current Electricity Chapter 2 Kirchhoff’s Laws And Electrical Measurement A voltage V is supplied to the potentiometer

Answer:

Equivalent resistance,

⇒ \(r=\frac{R_0}{2}+\frac{\frac{R_0}{2} \cdot R}{\frac{R_0}{2}+R}\)

⇒ \(=\frac{R_0}{2}+\frac{R_0 R}{R_0+2 R}=\frac{R_0^2+2 R_0 R+2 R_0 R}{2\left(R_0+2 R\right)}\)

⇒ \(\frac{R_0\left(R_0+4 R\right)}{2\left(R_0+2 R\right)}\)

Currently this circuit, \(I=\frac{V}{r}=\frac{2 V\left(R_0+2 R\right)}{R_0\left(R_0+4 R\right)}\)

So, the voltage across R is

⇒ \(V_R=V-I \cdot \frac{R_0}{2}=V-V \cdot \frac{R_0+2 R}{R_0+4 R}\)

⇒ \(\left(1-\frac{R_0+2 R}{R_0+4 R}\right) V=\frac{2 R}{R_0+4 R} V\)

Question 14. In the potentiometer circuit shown, the null point is at X State with reason, where the balance point will be shifted when

  1. Resistance R is increased, keeping all other parameters unchanged;
  2. Resistance S is increased, keeping R constant.

Class 12 Physics Unit 2 Current Electricity Chapter 2 Kirchhoff’s Laws And Electrical Measurement the potentiometer circuit

Answer:

1. When the resistance R is increased, the potential gradient increases resulting in a decrease in the balancinglength.

2. When the resistance S is increased, the terminal potential difference across the cell increases resulting in an increase in the balancing length.

Question 15. In a meter bridge, the balance point is found at a distance l1 with resistances R and S. An unknown resistance X is now connected in parallel to the resistance S, and the balance point is found at a distance l2. Obtain a formula for X in terms of l1, l2, and S.

Answer:

According to the problem,

⇒ \(\frac{R}{S}=\frac{l_1}{100-l_1}\)…(1)

When unknown resistance X is connected in parallel with resistance S, the effective resistance becomes \(\frac{S X}{S+X}\)

∴ we get \(\frac{R}{\left(\frac{S X}{S+X}\right)}=\frac{l_2}{100-l_2}\)…(2)

Dividing equation (1) by (2), we get

⇒ \(\frac{X}{S+X}=\frac{l_1}{l_2}\left(\frac{100-l_2}{100-l_1}\right)\)

or, \(l_2\left(100-l_1\right) X=l_1\left(100-l_2\right)(S+X)\)

or, \(\left\{l_2\left(100-l_1\right)-l_1\left(100-l_2\right)\right\} X=l_1\left(100-l_2\right) S\)

or, \(X=\frac{l_1\left(100-l_2\right)}{100\left(l_2-l_1\right)} S\)

where l1 and l2 are in cm

Question 16. The current is drawn from a cell of emf E and internal resistance r connected to the network of resistors each of resistance r. Obtain the expression for

  1. The Current Drawn From The Cell And
  2. The power consumed in the network.

Class 12 Physics Unit 2 Current Electricity Chapter 2 Kirchhoff’s Laws And Electrical Measurement the power consumed in the network

Answer:

The circuit can be redrawn as

Class 12 Physics Unit 2 Current Electricity Chapter 2 Kirchhoff’s Laws And Electrical Measurement Net resistance of the circuit

Between points A and B r, 2r, 2r, and r are in parallel.

So, \(\frac{1}{R_{A B}}=\frac{1}{r}+\frac{1}{r}+\frac{1}{2 r}+\frac{1}{2 r}\)

⇒ \(\frac{1}{R_{A B}}=\frac{3}{r} \quad \text { or, } R_{A B}=\frac{r}{3}\)

Netresistance of the circuit,

⇒ \(R=r+R_{A B}=r+\frac{r}{3}=\frac{4 r}{3}\)

1. Current drawn from the cell

⇒ \(I=\frac{E}{R}=\frac{E}{(4 r / 3)}=\frac{3 E}{4 r}\)

2. Power consumed in network,

⇒ \(P=I^2 R_{A B}\)

∴ \(P=\left(\frac{3 E}{4 r}\right)^2 \frac{r}{3}=\frac{3 E^2}{16 r}\)

Question 17. A resistance of R draws current from a potentiometer. The potentiometer wire AB, has a total resistance of R0. A voltage V is supplied to the potentiometer. Derive an expression for the voltage across R when the sliding contact is in the middle of the potentiometer wire.

Class 12 Physics Unit 2 Current Electricity Chapter 2 Kirchhoff’s Laws And Electrical Measurement potentiometer wire

Answer:

When the sliding contact is in the middle of the potentiometer, the total resistance between A and C is given by

Class 12 Physics Unit 2 Current Electricity Chapter 2 Kirchhoff’s Laws And Electrical Measurement middle of the potentiometer

⇒ \(\frac{1}{R_1}=\frac{1}{R}+\frac{1}{\left(R_0 / 2\right)} \quad \text { or, } R_1=\frac{R_0 R}{R_0+2 R}\)

The total resistance between A and B = Rl +R0/2

∴ The current flowing through the potentiometer,

⇒ \(I=\frac{V}{R_1+R_0 / 2}=\frac{2 V}{2 R_1+R_0}\)

The voltage V1 taken from the potentiometer is given by

⇒ \(V_1=I R_1=\left(\frac{2 V}{2 R_1+R_0}\right) \times R_1\)

⇒ \(\frac{2 V}{2\left(\frac{R_0 \times R}{R_0+2 R}\right)+R_0} \times \frac{R_0 \times R}{R_0+2 R}\)

= \(\frac{2 V R}{R_0+4 R}\)

Question 18. In a potentiometer arrangement for determining the emf of a cell, the balance point of the cell in an open circuit is 350 cm. When a resistance of 9Ω is used in the external circuit of the cell, the balance point shifts to 300 cm. Determine the internal resistance of the cell.
Answer:

Here, = 350 cm, l2 = 300 cm, R = 9Ω

The internal resistance of the cell,

⇒ \(r=\left(\frac{l_1}{l_2}-1\right) R=\left(\frac{350}{300}-1\right) \times 9=1.5 \Omega\)

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