A current of 2a flows through a 2 ohm resistor when connected across a battery , the same battery supplies a current of 0.5a when connected across a 9 ohm resistor . the internal resistance of the battery
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Let the internal resistance of the battery be r
Now,
The potential of the battery is constant or in other words the emf supplied by the battery will be same (No matter the amount of resistance connected)
So,
V1 = V2
2(r+ 2)= 0.5(r+ 9)
2r + 4 = 0.5 r + 4.5
1.5r = 0.5
r=1/3
r=0.33 ohm
So,
The internal resistance of the battery is 0.33ohm
Now,
The potential of the battery is constant or in other words the emf supplied by the battery will be same (No matter the amount of resistance connected)
So,
V1 = V2
2(r+ 2)= 0.5(r+ 9)
2r + 4 = 0.5 r + 4.5
1.5r = 0.5
r=1/3
r=0.33 ohm
So,
The internal resistance of the battery is 0.33ohm
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