WAEC - Physics (1988 - No. 36)
A cell can supply current of 0.8A and 0.4A through a 2Ω and a 3Ω resistors respectively. Calculate the internal resistance of the cell.
0.2Ω
0.4Ω
1.0Ω
3.0Ω
9.0Ω
Explanation
Assume the internal resistor = R
Current with 2 Ohm Resistor = 0.80A
Voltage = (R + 2) x 0.80 = V ; That is 0.8R + 1.6 = V ———————[[1]
Current with 3 Ohm Resistor = 0.40 A
Voltage = (R+3) x 0.40 = V ; That is 0.4R + 1.2 = V ————————[[2]
Subtracting 1 from 2 gives
--> (0.8 - 0.4) R
( 1.6 - 1.2 )
( V - V )
Therefore
0.4R + 0.4 = 0
R = 0.4 /0.4 = 1 Ohms
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