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A cell of e.m.f. 1.8V and internal resistance 2Ω is connected in series with an ammeter of resistance 0.7Ω and a resistor of 4.5Ω as shown in Fig. What would be the reading of the ammeter? - Physics

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Question

A cell of e.m.f. 1.8V and internal resistance 2Ω is connected in series with an ammeter of resistance 0.7Ω and a resistor of 4.5Ω as shown in Fig. 

  1. What would be the reading of the ammeter?
  2. What is the potential difference across the terminals of the cell? 
Numerical

Solution

The resistances are series connected

∴ Total resistance of circuit = (r + R)

(r + R) = (2 + 0.7 + 4.5)

= 7.2 Ω

e.m.f. of cell = 1.8 V

  1.  Reading of Ammeter `I = "V"/("R" + r)`
    = `("e.m.f")/("R" + r)`
    `I = 1.8/7.2`
    = 0.25 A
  2. Potential difference across the terminals of the cell = IR
    `"V" = 1/4 xx 5.2`
    V = 1.3 V
    R = 4.5 + 0.7
    R = 5.2 Ω
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Internal Resistance of a Cell
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Chapter 8: Current Electricity - Exercise 8 (B) 3 [Page 201]

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Selina Physics [English] Class 10 ICSE
Chapter 8 Current Electricity
Exercise 8 (B) 3 | Q 3 | Page 201

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