मराठी

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: a. The reading of the ammeter, - Physics

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प्रश्न

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:

  1. The reading of the ammeter,
  2. The potential difference across the terminals of the cells, and
  3. The potential difference across the 4.5 Ω resistor.
संख्यात्मक
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उत्तर

The total resistance of the circuit is

`"R"_"eq" = "R"_"cell" + "R"_"ammeter" + "R"_1 || "R"_2`

`therefore "R"_"eq" = 1.2 + 0.8 + ("R"_1"R"_2)/("R"_1 + "R"_2)`

`therefore "R"_"eq" = 2 + (4.5 xx 9)/(4.5 + 9)`

= `2 + 40.5/13.5`

`therefore "R"_"eq" = 5  Omega`

(a) Therefore, the current through the ammeter is

 I = `"E"_"cell"/"R"_"eq" = 2/5 = 0.4 "A"`

(b) The potential difference across the ends of the cells is

`"V"_"cell" = "E"_"cell" - "IR"_"cell"`

`therefore "V"_"cell" = 2 - 0.4 xx 1.2`

`therefore "V"_"cell" = 2 - 0.48 = 1.52 "V"`

(c) The potential difference across the 4.5 Ω resistor is

`"V"_4.5 = "V"_"cell" - "V"_"ammeter"` 

`therefore "V"_4.5 = 1.52 - 0.4 xx 0.8`

`therefore "V"_4.5 = 1.52 - 0.32`

`therefore "V"_4.5 = 1.2  "V"`

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  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 8: Current Electricity - EXERCISE - 8(B) [पृष्ठ २०३]

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सेलिना Physics [English] Class 10 ICSE
पाठ 8 Current Electricity
EXERCISE - 8(B) | Q 31. | पृष्ठ २०३

संबंधित प्रश्‍न

Name two factors on which the internal resistance of a cell depends and state how does it depend on the factors stated by you.


A battery of e.m.f 3.0 V supplies current through a circuit in which the resistance can be changed.
A high resistance voltmeter is connected across the battery. When the current is 1.5 A, the voltmeter reads 2.7 V. Find the internal resistance of the battery. 


A cell of emf. 1.5 V and internal resistance 10 ohms is connected to a resistor of 5 ohms, with an ammeter in series see fig.. What is the reading of the ammeter?


Four cells, each of e.m.f. 1.5 V and internal resistance 2.0 ohms are connected in parallel. The battery of cells is connected to an external resistance of 2.5 ohms. Calculate:

(i) The total resistance of the circuit.
(ii) The current flowing in the external circuit.
(iii) The drop in potential across-the terminals of the cells.


Define the e.m.f. (E) of a cell and the potential difference (V) of a resistor R in terms of the work done in moving a unit charge. State the relation between these two works and the work done in moving a unit charge through a cell connected across the resistor. Take the internal resistance of the cell as ‘r’. Hence obtain an expression for the current i in the circuit.


A battery of 4 cell, each of e.m.f. 1.5 volt and internal resistance 0.5 Ω is connected to three resistances as shown in the figure. Calculate:
(i) The total resistance of the circuit.
(ii) The current through the cell.
(iii) The current through each resistance.
(iv) The p.d. across each resistance.


A cell supplies a current of 0.6 A through a 2Ω coil and a current of 0.3 A through on 8Ω coil. Calculate the e.m.f and internal resistance of the cell.


(a) Calculate the total resistance across AB.

(b) If a cell of e.m.f 2.4 V with negligible internal resistance is connected across AB then calculate the current drawn from the cell.


Study the diagram:

  1. Calculate the total resistance of the circuit.
  2. Calculate the current drawn from the cell.
  3. State whether the current through 10 Ω resistor is greater than, less than or equal to the current through the 12 Ω resistor.

Explain the meaning of the term internal resistance of a cell.


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