Page No 131: - Chapter 3 - Current Electricity - Additional Exercise Solutions - Ncert Solutions class 12 - Physics

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Question 3.23:

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?

Answer:

Resistance of the standard resistor, R = 10.0 Ω

Balance point for this resistance, l₁ = 58.3 cm

Current in the potentiometer wire = i

Hence, potential drop across R, E₁ = iR

Resistance of the unknown resistor = X

Balance point for this resistor, l₂ = 68.5 cm

Hence, potential drop across X, E₂ = iX

The relation connecting emf and balance point is,

Therefore, the value of the unknown resistance, X, is 11.75 Ω.

If we fail to find a balance point with the given cell of emf, ε, then the potential drop across R and X must be reduced by putting a resistance in series with it. Only if the potential drop across R or X is smaller than the potential drop across the potentiometer wire AB, a balance point is obtained.

Question 3.24:

Figure 3.35 shows a 2.0 V potentiometer used for the determination of internal resistance of a 1.5 V cell. The balance point of the cell in open circuit is 76.3 cm. When a resistor of 9.5 Ω is used in the external circuit of the cell, the balance point shifts to 64.8 cm length of the potentiometer wire. Determine the internal resistance of the cell.

Answer:

Internal resistance of the cell = r

Balance point of the cell in open circuit, l₁ = 76.3 cm

An external resistance (R) is connected to the circuit with R = 9.5 Ω

New balance point of the circuit, l₂ = 64.8 cm

Current flowing through the circuit = I

The relation connecting resistance and emf is,

Therefore, the internal resistance of the cell is 1.68Ω.