Question

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

(i) minimum current flowing 

(ii) maximum current flowing 

(a) How will you connect the resistances in each case ? . 

(b) Calculate the strength of the total current in the circuit in the two cases.

Answer 1

(a)

(i) For minimum current we must make R maximum. This can be done by connecting the resistances in series.

(ii) For maximum current we must make R minimum. This can be done by connecting the resistances in parallel.

(b)

(i) \(R_{eq} = R_1 + R_2 \)

\(= 5 + 10 \)

\(= 15\Omega\)

\(I = \frac V{R_{eq}}\)

\(= \frac 6{15}\)

\(= 0.4A\)

(ii) \(R_{eq} = \frac{R_1R_2}{R_1 + R_2}\)

\(= \frac{5 \times 10}{5 + 10}\)

\(= 3.33 \Omega\)

\(I = \frac V{R_{eq}}\)

\(= \frac 6{3.33}\)

\(= 1.8A\)

Answer 2

Given: Two resistors with resistances R1=5ohm and R2=10ohm, V=6volt 

(a) For minimum current these two should be connected in series. For maximum current these two should be connected in parallel. 

(b)In series, 

Total resistance = 5+10 = 15ohms 

Therefore total current drawn = V/R = 6/15 = 0.4amps 

In parallel, 

Total resistance R is given as 1/R=1/R₁+1/R₂ 

1/R =1/5+1/10 

1/R=3/10 

R=10/3 ohm 

Therefore total current drawn by the circuit = V/R = 6/(10/3) =1.8amps.