Two resistors, with resistances, 5Ω and 10Ω respectively are to be connected to a battery of emf 6V so as to obtain:

Since, V=IR, therefore,(a) to obtain

(i) minimum current flowing, the R_net should be higher so, it should be connected in series.

(ii)maximum current flowing, the R_net should be lower so it should be connected in parallel.

(b)

i. The strength of total current in minimum current flowing circuit is,

R_net = 5Ω + 10Ω = 15 Ω

Since,  V=IR

6 V = I 15Ω

I = 0.4 A

ii. The strength of total current in maximum current flowing circuit is,

1/R_net = 1/R₁ +1/R_2

1/R_net = 1/5 + 1/10

R_net = 3.33 Ω

Since,

V = IR

6V = I 3.33Ω

I = 1.801 A

The strength of total current in minimum flowing circuit is 0.4 A and in maximum current flowing is 1.801 A.