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Ashok0210
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Figure 3.34 shows a potentiometer circuit for comparison of two resistances. The balance point with a standard resistor R = 10.0 Ω is found to be 58.3 cm, while that with the unknown resistance X is 68.5 cm. Determine the value of X. What might you do if you failed to find a balance point with the given cell of emf ε ?

Class 12 Physics CBSE and UP Board Current Electricity Additional Exercise Chapter-3 Exercise 3.23 NCERT Solutions Class 12 Physics Chapter 3 Question 23

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Ashok0210
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Figure 3.33 shows a potentiometer with a cell of 2.0 V and internal resistance 0.40 Ω maintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant emf of 1.02 V (for very moderate currents upto a few mA) 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 ε and the balance point found similarly, turns out to be at 82.3 cm length of the wire.(a) What is the value ε? (b) What purpose does the high resistance of 600 kΩ have?(c) Is the balance point affected by this high resistance? (d) Is the balance point affected by the internal resistance of the driver cell? (e) Would the method work in the above situation if the driver cell of the potentiometer had an emf of 1.0V instead of 2.0V? (f ) Would the circuit work well for determining an extremely small emf, say of the order of a few mV (such as the typical emf of a thermo-couple)? If not, how will you modify the circuit?

Figure 3.33 shows a potentiometer with a cell of 2.0 V and internal resistance 0.40 Ω maintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant emf of 1.02 V (for very moderate currents upto a few mA) 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 ε and the balance point found similarly, turns out to be at 82.3 cm length of the wire.(a) What is the value ε? (b) What purpose does the high resistance of 600 kΩ have?(c) Is the balance point affected by this high resistance? (d) Is the balance point affected by the internal resistance of the driver cell? (e) Would the method work in the above situation if the driver cell of the potentiometer had an emf of 1.0V instead of 2.0V? (f ) Would the circuit work well for determining an extremely small emf, say of the order of a few mV (such as the typical emf of a thermo-couple)? If not, how will you modify the circuit?

Class 12 Physics CBSE and UP Board Current Electricity Additional Exercise Chapter-3 Exercise 3.22 NCERT Solutions Class 12 Physics Chapter 3 Question 22

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Ashok0210
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Determine the current drawn from a 12V supply with internal resistance 0.5Ω by the infinite network shown in Fig. 3.32. Each resistor has 1Ω resistance.

Class 12 Physics CBSE and UP Board Current Electricity Additional Exercise Chapter-3 Exercise 3.32 NCERT Solutions Class 12 Physics Chapter 3 Question 32

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Ashok0210
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(a) Given n resistors each of resistance R, how will you combine them to get the (i) maximum (ii) minimum effective resistance? What is the ratio of the maximum to minimum resistance? (b) Given the resistances of 1 Ω, 2 Ω, 3 Ω, how will be combine them to get an equivalent resistance of (i) (11/3) Ω (ii) (11/5) Ω, (iii) 6 Ω, (iv) (6/11) Ω? (c) Determine the equivalent resistance of networks shown in Fig. 3.31.

Class 12 Physics CBSE and UP Board Current Electricity Additional Exercise Chapter-3 Exercise 3.20 NCERT Solutions Class 12 Physics Chapter 3 Question 20

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Class 12 Physics CBSE and UP Board Current Electricity Additional Exercise Chapter-3 Exercise 3.19 NCERT Solutions Class 12 Physics Chapter 3 Question 19

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Class 12 Physics CBSE and UP Board Current Electricity Additional Exercise Chapter-3 Exercise 3.18 NCERT Solutions Class 12 Physics Chapter 3 Question 18

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Class 12 Physics CBSE and UP Board Current Electricity Additional Exercise Chapter-3 Exercise 3.16 NCERT Solutions Class 12 Physics Chapter 3 Question 16

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Class 12 Physics CBSE and UP Board Current Electricity Additional Exercise Chapter-3 Exercise 3.15 NCERT Solutions Class 12 Physics Chapter 3 Question 15

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Class 12 Physics CBSE and UP Board Current Electricity Additional Exercise Chapter-3 Exercise 3.14 NCERT Solutions Class 12 Physics Chapter 3 Question 14

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