a cell of emf E nd internal resistance R is connected to 2 external resistances R1 n R2 nd a perfect ammeter ..the current in the circuit is measured in 4 difffernt conditions .. 1) without any external resistance 2) with R1 only 3) with R1 n R2 in series 4) with R1 n R2 in parallel the current measuere in 4 cases are 0.4 A ,1.05A , 1.4 A nd 4.2 IDENTIFY THE CURRENT CORRESPONDING TO THE 4 CASES MENTIONED ABOVE

Asked by priya mandal | 8th Jun, 2013, 07:51: PM

Expert Answer:

Please note that your question is not very clear. You are first asking to calculate the current in the 4 cases and then, you have yourself given the values of current in 4 cases. 
You can find below how to find current in four cases. 
Using Ohms law V = IR for all the cases. In this case V = E
1. In the absence of external resistor: I = E/R
2. With R1 only, then the equivalent resistance in circuit Req = R+R1
Hence current I = E/(R+R1)
3. with R1 and R2 in series, then the equivalent resistance in circuit Req = R+ R1+R2
Hence current I = E/(R+R1+R2)
4. With R1 and R2 in parallel. Hence their equivalent resistance R' will be given by 
1/R' = 1/R1+1/R2 
R' = R1R2/(R1+R2)
Now this is connected in series with internal resistance of cell, hence R eq = R' + R = R+R1R2/(R1+R2)
Hence, I = E/ [R+R1R2/(R1+R2)]

Answered by  | 9th Jun, 2013, 06:30: AM

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