Two resistances of 6 KΩ and 12 kΩ are connected in parallel. A 6V battery is connected across its ends, find the values of the following quantities:

(a) Equivalent resistance of the parallel combination. 
(b) Current passing through each of the resistances. 
(c) Potential difference across each of the resistance. 
 Ans. [(a) 4 kΩ, (b) 1 mA,0.5 mA (c) 6 V]



Data Given:


Resistance 1 = R₁ = 6 KΩ = 6 ｘ10³ Ω

Resistance 2 = R₂ = 12 KΩ = 12 ｘ10³ Ω

Voltage = V = 6 V



To Find:


(a) Equivalent Resistance = Rₑ = ?  

(b) Potential difference through resistance R₁ = V₁ = ?
    Potential difference through resistance R₂ = V₂ = ?

(c) Current passing through resistance R₁ = I₁ = ? 
     Current passing through resistance R₂ = I₂ = ?



Solution:


(a) Equivalent Resistance = Rₑ = ?  


We have the formula Equivalent Resistance connected in Parallel as 

`\frac{1}{Rₑ}` = `\frac{1}{R₁}` + `\frac{1}{R₂}`

Now by putting the values

`\frac{1}{Rₑ}` = `\frac{1}{6 ｘ 10³ Ω}` + `\frac{1}{12 ｘ 10³ Ω}`

`\frac{1}{Rₑ}` = `\frac{2 + 1}{12 ｘ 10³}` Ω

`\frac{1}{Rₑ}` = `\frac{3}{12 ｘ 10³}` Ω

or (flipping the fraction doth side)

Rₑ = `\frac{12 ｘ 10³}{3}` Ω

Rₑ = 4 𝐱 10³ F = 4 KΩ--------------Ans.1



(b) Potential difference through resistance R₁ = V₁ = ?
    Potential difference through resistance R₂ = V₂ = ? 


Potential Difference (voltage) through each resistance have the same (equal) values in a parallel combination of resistors So,

V = V₁ = V₂ = 6 V -------------Ans.2



(c) Current passing through resistance R₁ = I₁ = ? 
     Current passing through resistance R₂ = I₂ = ?

 
Current passing through resistance R₁
By using formula

V₁ = I₁ R₁

or 

I = `\frac {V₁}{R₁}`

I = `\frac {6 V}{6ｘ10³ Ω}`

I = 1 ｘ 10⁻³A

or

I = 1 mA -------------Ans. 3

Current passing through resistance R₂
By using formula
V₂ = I₂ R₂
or 

I = `\frac{V₂}{R₂}`

I = `\frac{6 V}{12ｘ10³ Ω}`

I = 0.5 ｘ 10⁻³ A

or

I = 0.5 mA ---------------Ans. 4

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