Three resistors of resistances R_1, R_2 and R_3 are connected in (i) series and (ii) parallel. Write expression for the equivalent resistance of the combination in each case. (b) Two identical resistance of 12 Omega each are connected to a battery of 3V. Calculate the ratio of the power consumed by the resulting combinations with minimum resistance and maximum resistance.

Three resistors of resistances R_1, R_2 and R_3 are connected in (i) s

1.	Three resistors of resistances R_1, R_2 and R_3 are connected in (i) series and (ii) parallel. Write expression for the equivalent resistance of the combination in each case. (b) Two identical resistance of 12 Omega each are connected to a battery of 3V. Calculate the ratio of the power consumed by the resulting combinations with minimum resistance and maximum resistance.
Answer» Solution :(i) In SERIES arrangement equivalent resistance `R_s=R_1+R_2+R_3` (ii) In parallel arrangement equivalent resistance `R_p` is GIVEN as: `1/R_p=1/R_1+1/R_2+1/R_3` (B) Here `R_1=R_2=12 Omega and V=3V` For minimum resistance, two RESISTORS must be connected in parallel so that `1/R_p=1/R_1+1/R_2=1/12+1/12=1/6 implies R_p=6 Omega` Hence power `P_p=V^2/R_p=(3)^2/6=1.5W` For maximum resistance two resistors must be connected in series so that `R_s=R_1+R_2=12+12=24 Omega` So the power, `P_s=V^2/R_s=(3)^2/24=0.375W implies P_p/P_s=1.5/0.375=4/1`