Question

Two resistors with resistances 50 and 10 Q are to be connected to a battery of emf 6 V so as to obtain:
(i) minimum current
(ü) maximum current
(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

SolutioN :

• R1 = 50Ω.
• R2 = 10Ω.

Condition 1.

• ✎ When we connect 50Ω and 10Ω in Parellel Combination.

$\tt \dagger \: \: \: \: \: \dfrac{1}{R_P}= \dfrac{1}{R_1}+ \dfrac{1}{R_2}+... \dfrac{1}{R_n}$

$\tt : \implies \dfrac{1}{R_P}= \dfrac{1}{50}+ \dfrac{1}{10}$

$\tt : \implies \dfrac{1}{R_P}= \dfrac{1 + 5}{50}$

$\tt : \implies \dfrac{1}{R_P}= \dfrac{6}{50}$

$\tt : \implies {R_P}= \dfrac{50}{6}\, \Omega.$

$\rule{200}3$

Condition 2.

• ✎ When we connect 50Ω and 10Ω in Series Combination.

$\tt \dagger \: \: \: \: \: {R_s}= {R_1}+{R_2}+...{R_n}$

$\tt : \implies{R_s}= 10 + 50 \, \Omega.$

$\tt : \implies{R_s}= 60 \, \Omega.$

Let's Find, Current.

• Current ( I )
• We know, Ohm' s Law.
• V = IR.
• Voltage ( V )
• Resistance ( R )

☛ Case 1.

• Rp = 50 / 6.

$\tt : \implies V = IR.$

$\tt : \implies 6 = I \times \dfrac{50}{6} .$

$\tt : \implies 6 \times \dfrac{6}{50} = I$

$\tt : \implies I = \dfrac{36}{50} .$

$\tt : \implies I = 0.72\, A .$

$\rule{200}1$

☛ Case 2.

$\tt : \implies V = IR.$

$\tt : \implies 6= I \times 60.$

$\tt : \implies \dfrac{1}{10} = I$

$\tt : \implies I = 0.1\,A .$

$\rule{200}3$

• ( i ) When we connect 50Ω and 10Ω in parellel Combination.( Maximum Current )
• ( ii ) When we connect 50Ω and 10Ω in Series Combination.( Minimum Current )
• ( a ) Series and parellel
• ( b ) I = 0.72 A.

Answer 2

SolutioN :

R1 = 50Ω.

R2 = 10Ω.

Condition 1.

✎ When we connect 50Ω and 10Ω in Parellel Combination.