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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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प्रश्न

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? 
संख्यात्मक
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उत्तर

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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अध्याय 8: Current Electricity - EXERCISE - 8(B) [पृष्ठ २०१]

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सेलिना Concise Physics [English] Class 10 ICSE
अध्याय 8 Current Electricity
EXERCISE - 8(B) | Q 3. | पृष्ठ २०१

संबंधित प्रश्न

A cell of Emf 2 V and internal resistance 1.2 Ω is connected with an ammeter of resistance 0.8 Ω and two resistors of 4.5 Ω and 9 Ω as shown in the diagram below:

1) What would be the reading on the Ammeter?

2) What is the potential difference across the terminals of the cell?


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A battery of e.m.f 3.0 V supplies current through a circuit in which the resistance can be changed.
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A cell of e.m.f. 2 V and internal resistance 1.2 Ω is connected to an ammeter of resistance 0.8 Ω and two resistors of 4.5 Ω and 9 Ω as shown in following figure.

Find:

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  2. The potential difference across the terminals of the cells, and
  3. The potential difference across the 4.5 Ω resistor.

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