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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.
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 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.
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 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.
A battery of e.m.f. 6.0 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 3 A, the voltmeter reads 5.4 V. Find the internal resistance of the battery.
