Two resistors with resistance = 5 ohm and 10 ohm respectively are to be connected to a battery of 6v

Q. 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

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

Ans. (i) (a) For obtaining minimum current, the two resistors should be connected in parallel.

(b) For obtaining maximum current, the two resistors should be connected in series.

= 1/5 + 1/10 = 3/10

====> R = 10/3 Ω

Total current in the circuit,

I = V /R = 6×3/10 = 1.8 A

(iii) For series combination,

R =  5 + 10 = 15 Ω

Total current in the circuit,

I = V /R =6/15 = 0.4 A

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:  

Resistance in parallrl arrangement is given 

`1/R=1/R_1+1/R_2` 

Here `R_1=5`Ω
         `R_2=10`Ω 

`1/R=1/5+1/10` 

`or1/R=(10+5)/50`  

`or1/R=15//50` 

`R=50/15=3.33`ΩTotal resistance = 3.33 Ω

Therefore, strength of the total current, I = V / R

Therefore, strength of the total current, I = 6 / 15 = 0.4 A 

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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(b) Effective resistance for parallel connection:

$\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}$
$\frac{1}{R}=\frac{R_1+R_2}{R_1\times R_2}$
$\therefore R=\frac{R_1\times R_2}{R1+R2}$
$=\frac{5\times 10}{5+10}$
$=\frac{50}{15}$
$=\frac{10}{3}\Omega$

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

(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 $I_s=\frac{V}{R}$
$=\frac{6}{15}$
$=0.4A$

$\therefore$total current when connected in series $I_s=0.4Ampere$

The total current when connected in parallel $I_p=\frac{V}{R}$
$=\frac{(6\times 3)}{10}$
$=1.8Ampere$
$\therefore$ total current when connected in parallel $I_p=1.8Ampere$