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:
(a) minimum current flowing
(b) maximum current flowing
(c) How will you connect the resistances in each case?
(d) Calculate the strength of the total current in the circuit in the two cases.

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Solution

Given: $R_{1} = 5 Ω, R_{2} = 10 Ω$ , EMF = 6V

a) Effective resistance for series connection: $R = R_{1} + R_{2}$
$R = 5 + 10$
$R = 15 Ω$
By Ohm's law, we have V=IR
so, $I_{m i n} = \frac{V}{R}$
$= \frac{6}{15}$
$= 0.4$
$∴ I_{m i n} = 0.4 A$

(b) Effective resistance for parallel connection:

$\frac{1}{R} = \frac{1}{R_{1}} + \frac{1}{R_{2}}$
$\frac{1}{R} = \frac{R_{1} + R_{2}}{R_{1} \times R_{2}}$
$∴ R = \frac{R_{1} \times R_{2}}{R 1 + R 2}$
$= \frac{5 \times 10}{5 + 10}$
$= \frac{50}{15}$
$= \frac{10}{3} Ω$

By Ohm's law, we have $V = I R$
so, $I = \frac{V}{R}$
$\Rightarrow (\frac{6}{10}) \times 3 = 1.8 A$
$I_{m a x} = 1.8 A$

(c) To get minimum current, the two resistors should be connected in series.

To get maximum current, we should connect the resistance in parallel

(d) The total current when connected in series $I_{s} = \frac{V}{R}$
$= \frac{6}{15}$
$= 0.4 A$

$∴$ total current when connected in series $I_{s} = 0.4 A m p e r e$

The total current when connected in parallel $I_{p} = \frac{V}{R}$
$= \frac{(6 \times 3)}{10}$
$= 1.8 A m p e r e$
$∴$ total current when connected in parallel $I_{p} = 1.8 A m p e r e$

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