Question

A 12V battery is connected to a circuit having three resisters of resistances 2Ω , 3Ω and

6 Ω in parallel. Calculate the total resistance of a circuit and find the current passing

through each resister.​

Answer 1

$\sf{\huge{\underline{\green{Given :-}}}}$

• A 12V battery is connected to a circuit having three resisters of resistances 2Ω , 3Ω and 6 Ω in parallel.

$\sf{\huge{\underline{\green{To\:Find :-}}}}$

• The total resistance of a circuit .

• The current passing .

$\sf{\huge{\underline{\green{Answer :-}}}}$

According to the question,

• V = 12 V

• R1 = 2Ω

• R2 = 3Ω

• R3 = 6Ω

Case 1 :-

Total resistance of parallel = $\dfrac{1}{Rp}$

➝ $\dfrac{1}{Rp}$ = $\dfrac{1}{R1}$ + $\dfrac{1}{R2}$ + $\dfrac{1}{R3}$

➝ $\dfrac{1}{Rp}$ = $\dfrac{1}{2}$ + $\dfrac{1}{3}$ + $\dfrac{1}{6}$

➝ $\dfrac{1}{Rp}$ = $\dfrac{3 + 2 + 1}{6}$

➝ $\dfrac{1}{Rp}$ = $\cancel{\dfrac{6}{6}}$

➝ $\dfrac{1}{Rp}$ = 1

The Total resistance of parallel is 1Ω .

Case 2 :-

We know that,

V = I R

➝ I = $\dfrac{V}{R}$

➝ I = $\dfrac{12}{1}$

➝ I = 12 A

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