the current in a simple series circuit is 5 ampere when an additional resistance of 2 ohm is inserted the current drops to 4 ampere the orignal resistance of circuit is once tell please though its a simple qsn
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Answered by
39
I = 5 A Resistance = R
I' = 4 A R' = R+2 Ω = total resistance
We assume R is the total effective resistance in the circuit in series with the battery. Also, that 2 Ω resistance is added in series with the= given resistance and battery.
Voltage of the battery = E = I R = I' R'
5 * R = 4 ( R +2)
=> R = 8 Ω
battery potential is = 40 V
I' = 4 A R' = R+2 Ω = total resistance
We assume R is the total effective resistance in the circuit in series with the battery. Also, that 2 Ω resistance is added in series with the= given resistance and battery.
Voltage of the battery = E = I R = I' R'
5 * R = 4 ( R +2)
=> R = 8 Ω
battery potential is = 40 V
kvnmurty:
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Answered by
21
First pls type the complete question , In complete question , it has been said that V is same , so ,
Ohk then let R is the original resistance then ,
V/R = 5 & V/R+2 = 2
Dividing ,
R+2/ R = 5/4
R = 8 ohm
Ohk then let R is the original resistance then ,
V/R = 5 & V/R+2 = 2
Dividing ,
R+2/ R = 5/4
R = 8 ohm
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