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प्रश्न
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.
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उत्तर
Let total e.m.f. of 4 cells = nE (n = number of cells)
E = 4 × 1.5
E = 6 volts ....(i)
Total internal resistance = nr (n = 4, r = 0.5 Ω)
Total internal resistance = 4 × 0.5 (in series)
= 2 Ω ....(ii)
Let total external resistance = X Ω
`1/"R" = 1/"R"_1 + 1/"R"_2` (R1 and R2 are in parallel)
`1/"R" = 1/4 + 1/12 = (3 + 1)/12 = 4/12`
R = `12/4 = 3 Ω`
Total external resistance = X = (R + R3)Ω (in series)
X = 3Ω + 7 Ω = 10Ω .....(iii)
Total resistance of the circuit = X + r = 10 + 2 = 12 Ω
Total current through the cell = `"E"/("X" + "r")` ...(as E = I (R + r))
I = `6/(10 + 2) = 0.5` A ....from (i), (ii) and (iii)
Current through resistor 4 Ω =
I1 = `("I" xx "R"_2)/("R"_1 + "R"_2)`
I1 = `(0.5 xx 12)/(4 + 12) = (0.5 xx 12)/16 = (0.5 xx 3)/4`
I1 = `1.5 xx 1/4 = 0.375`A
Similarly current through resistor 12 Ω = I2
I2 = `"I" xx "R"_1/("R"_1 + "R"_2)`
I2 = `0.5 xx 4/16 = 0.5 xx 1/4 = 1/8`
I2 = 0.125 A.
P.D. across resistance 7 Ω = V = I × R = 0.5 × 7 = 3.5 V
P.D. across resistance 4 Ω = V1 = V1 × R1 = 0.375 × 4 = 1.5 V
P.D. across resistance 12 Ω = V2 = I2 × R2 = 0.125 × 12 = 1.5 V.
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?
Explain why the p.d across the terminals of a cell is more in an open circuit and reduced in a closed circuit.
The diagram below in Fig. 8.40 shows a cell of e.m.f. ε = 2 volt and internal resistance r = 1 ohm to an external resistance R = 4 ohm. The ammeter A measures the current in the circuit and the
voltmeter V measures the terminal voltage across the cell. What will be the readings of the ammeter and voltmeter when (i) the key K is open, (ii) the key K is closed.
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 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.
What is the colour code for the insulation on the earth wire?
Four cells each of e.m.f. 2V and internal resistance 0.1 Ω are connected in series to an ammeter of negligible resistance, a 1.6 Ω resistor and an unknown resistor R1. The current in the circuit is 2A. Draw a labelled diagram and calculate:
(i) Total resistance of the circuit,
(ii) Total e.m.f.
(iii) The value of R1 and
(iv) The p.d. across R1.
When a resistance of 3Ω is connected across a cell, the current flowing is 0.5 A. On changing the resistance to 7Ω, the current becomes 0.25A. Calculate the e.m.f. and the internal resistance of the cell.
Explain the meaning of the term internal resistance of a cell.
The diagram in Figure shows a cell of e.m.f. ε = 4 volt and internal resistance r = 2 ohm connected to an external resistance R = 8 ohm. The ammeter A measures the current in the circuit and the voltmeter V measures the terminal voltage across the cell. What will be the readings of the ammeter and voltmeter when
- the key K is open, and
- the key K is closed
