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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:
- The reading of the ammeter,
- The potential difference across the terminals of the cells, and
- The potential difference across the 4.5 Ω resistor.
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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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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?
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 cell of e.m.f ε and internal resistance r is used to send current to an external resistance R. Write expressions for
- the total resistance of circuit.
- the current drawn from the cell.
- the p.d. across the cell.
- voltage drop inside the cell.
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 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?
- What is the potential difference across the terminals of the cell?
A battery of e.m.f. 15 V and internal resistance 3 ohm is connected to two resistors of resistances 3 ohm and 6 ohm is series Find:
(a) the current through the battery
(b) the p.d. between the terminals 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?
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) 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:
- Calculate the total resistance of the circuit.
- Calculate the current drawn from the cell.
- State whether the current through 10 Ω resistor is greater than, less than or equal to the current through the 12 Ω resistor.
