Q. 5. A battery made of 5 cells each of 2V and have internal resistance 0.12.0.2 2,032,
0.42 and 0.5 Q is connected across 10 ohm resistance. Draw circuit resistance. Draw
circuit diagram and calculate the current flowing through 10 resistance.
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Answer:
Given,
That, 5 cells having internal resistance of 0.1 Ω , 0.2 Ω , 0.3 Ω , 0.4 Ω , 0.5 and a resistance of 10 Ω are connected in the series.
Hence,
Total resistance of the circuit,
R = 0.1 + 0.2 + 0.3 + 0.4 + 0.5 + 10
R = 1.5 + 10
R = 11.5 Ω
Given that 5 cells are connected in the series, Applying the second Kirchhoff's in the closed loop, Total voltage,
V = 2 + 2 + 2 + 2 + 2
V = 10 volts.
Using ohm's law,
Current in circuit,
I = V/R
I = 10/11.5
I = 0.869 A
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