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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.

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