Two resistances of 2 kΩ and 8 kΩ are joined in series, if a 10 V battery is connected across the ends of this combination, find the following quantities:
(a) The equivalent resistance of the series combination. 
(b) Current passing through each of the resistances. 
(c) The potential difference across each resistance. 
 Ans. [(a) 10 kΩ (b) 1 mA (c) 2 V, 8 V]



Data Given:


Resistance 1 = R₁ = 2 KΩ = 2 ｘ10³ Ω

Resistance 2 = R₂ = 8 KΩ = 8 ｘ10³ Ω

Voltage = V = 10 V


To Find:


(a) Equivalent Resistance = Rₑ = ?  

(b) Current passing through each resistance = I = ? 

(c) Potential difference through resistance 1 = V₁ = ?
    Potential difference through resistance 2 = V₂ = ?



Solution:


(a) Equivalent Resistance = Rₑ = ?  


In series combination

Equivalent Resistance = Rₑ = R₁ + R₂

by putting values 

Rₑ = 2 ｘ10³ Ω + 8 ｘ10³ Ω

Rₑ = 10ｘ10³ Ω 

Rₑ = 10 KΩ  -----------------Ans.



(b) Current passing through each resistance = I = ?

 
Current passing through each resistance has the same (equal) values in the series combination of resistors So,

I = I₁ = I₂

Now by using the formula

V = I Rₑ

or 

I = `\frac{V}{Rₑ}`

I = `\frac{10 V}{10ｘ10³ Ω}`

I = 1 ｘ 10⁻³A

or

I = 1 mA -------------------Ans.2




(c) Potential difference through resistance 1 = V₁ = ?
    Potential difference through resistance 2 = V₂ = ?

 
As the potential difference (voltage) is different depends on the value of the resistance 

So, the Potential difference (voltage) are across resistance 1 will be  

V₁ = I R₁

by putting values

V₁ = (1 ｘ 10⁻³A) (2 ｘ10³ Ω)

V₁ = 2 ｘ 10⁻³⁺³ V

V₁ = 2 V -------------------Ans. 3



And the Potential difference (voltage) across the resistance 2 are 

V₂ = I R₂

by putting values

V₂ = (1 ｘ 10⁻³A) (8 ｘ10³ Ω)

V₂ = 8 ｘ 10⁻³⁺³ V

V₂ = 8 V -----------------Ans.4

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