Question

calculate the equivalent resistance when two resistance of 8ohms and 12ohms are connected in parallel​

Answer 1

Given :

Two resistor are connected in parallel

• R₁ = 8Ω
• R₂ = 12Ω

To find :

The equivalent resistance.

Solution :

The reciprocal of the combined resistance of a number of resistance connected in parallel is equal to the sum of the reciprocal of all the individual resistances. .i.e.,

${ \leadsto{ \bf{\dfrac{1}{R} = \dfrac{1}{R_1} + \dfrac{1}{R_2}}}}$

${ \leadsto{ \sf{\dfrac{1}{R} = \dfrac{1}{8} + \dfrac{1}{12}}}}$

${ \leadsto{ \sf{\dfrac{1}{R} = \dfrac{3 + 2}{24}}}}$

${ \leadsto{ \sf{\dfrac{1}{R} = \dfrac{5}{24}}}}$

${ \leadsto{ \sf{R = \dfrac{24}{5}}}}$

${ \leadsto{ \sf{R = 4.8 \: \Omega}}}$

thus, the equivalent resistance of 8Ω and 12Ω is 4.8Ω

Remember !

SI unit of resistance is ohms (Ω)

Answer 2

$\huge\underline{\bf \orange{Aηsωer :}}$

• The reciprocal of the combined resistance of a number of resistance connected in parallel is equal to the sum of the reciprocal of all the individual resistances. .i.e.,

⇝ 1/R = 1/RR + 2/R1

⇝ 1/R = 1/8 + 1/12

⇝ 1/R= 3+2/24

⇝ 1/R = 5/24

⇝R= 24/5

⇝R= 4.8Ømega Ans.

Thus, the equivalent resistance of 8Ω and 12Ω is 4.8Ω