Two resistors with resistances 5 ohm and 10 ohm are to be connected to a battery of emf 6 V so as to obtain :
1) minimum current.
2) maximum current.
A) How will you connect the resistance in each case .
B) Calculate the strength of total current in the circuit in the 2 cases.

2

Answers

2014-05-29T21:48:39+05:30
When the resistant is connected in series currenr=6/10 amp abd in parrallel current=6/10/3=18/10=9/5 amp
4 3 4
2014-05-30T12:42:01+05:30
A)  CASE 1:- If connected in series:
Potential Difference(V) = 6V, R1 = 5 Ohm, R2 = 10 Ohms
Therefore, Equivalent resistance(R) = R1 + R2 = 5 + 10 = 15 OHMS
V = IR                          (Ohm's Law)
6 = I * 15
I = 6/15 = 0.4 Amperes                  -----------(i)

CASE 2 :- If connected in parallel:
Potential Difference(V) = 6V, R1 = 5 Ohm, R2 = 10 Ohms
Therefore, Equivalent resistance(R) = 1/R1 + 1/R2 = 1/5 + 1/10 = 0.3 Ohms
V = IR                          (Ohm's Law)
6 = I * 0.3
I = 20 Amperes                            -------------(ii)
Therefore as eqn i > ii. So to get minimum current we will arrange them in series and to get maximum current we will arrange them in parallel



2 5 2