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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 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.
Explain why the p.d across the terminals of a cell is more in an open circuit and reduced in a closed circuit.
A cell is used to send current to an external circuit.
- How does the voltage across its terminals compare with its e.m.f.?
- Under what condition is the e.m.f. of a cell equal to its terminal voltage?
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 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.
What is the colour code for the insulation on the earth wire?
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.
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.
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

