Question

Two resistors, with resistance 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) For the minimum current flowing in the circuit, the resistors should be connected in series and for the maximum current in the circuit, the resistors should be connected in parallel with the battery.
(b) When the resistors are connected in parallel:

$$\begin{gathered} \mathrm{Resistance}\mathrm{in}\mathrm{parallel}\mathrm{arrangement}\mathrm{is}\mathrm{given} \\ \frac{1}{R}=\frac{1}{R_{\mathit{1}}}+\frac{1}{R_{\mathit{2}}} \\ \mathrm{Here}{R}_1=5\mathrm{\Omega } \\ \mathit{}{R}_{\mathit{2}}=10\mathrm{\Omega } \\ \frac{1}{R}=\frac{1}{5}+\frac{1}{10} \\ \mathrm{or}\frac{1}{R}=\frac{10+5}{50} \\ \mathrm{or}\frac{1}{R}=\frac{15}{50} \\ \mathit{}R=\frac{50}{15}=3.33\mathrm{\Omega } \end{gathered}$$
Total resistance = 3.33 Ω
Therefore, strength of the total current, I = V / R
I = 6 / 3.33
I = 1.8 A
When the resistors are connected in series, the resultant resistance is given by R = R₁ + R₂
Here, R₁ = 5 $\mathrm{\Omega }$
R₂= 10 $\mathrm{\Omega }$
So, R = 5 $\mathrm{\Omega }$ + 10 $\mathrm{\Omega }$ = 15 $\mathrm{\Omega }$
Total resistance = 15 Ω
Therefore, strength of the total current, I = 6 / 15 = 0.4 A