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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?

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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
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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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Chapter 8: Current Electricity - EXERCISE - 8(B) [Page 201]

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

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