Question

Three resistance 4ohm!,6ohm, and 12ohm, are connected in parallel with source of 6volt. find the current flowing through the circuit.​

Answer 1

Given :

Three resistors are connected in parallel,

• R₁ = 4 ohms
• R₂ = 6 ohms
• R₃ = 12 ohms

Potential difference : 6 volt

To find :

The current flowing through the circuit.

Solution :

first we have to find the total resistance of the circuit.

We know that,

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/R₁ + 1/R₂

By substituting the values in the formula,

$\dashrightarrow \sf \dfrac{1}{R_{eq}} = \dfrac{1}{R_{1}} + \dfrac{1}{ R_2} + \dfrac{1}{ R_3}$

$\dashrightarrow \sf \dfrac{1}{R_{eq}} = \dfrac{1}{4} + \dfrac{1}{6} + \dfrac{1}{12}$

$\dashrightarrow \sf \dfrac{1}{R_{eq}} = \dfrac{3 + 2 + 1}{12}$

$\dashrightarrow \sf \dfrac{1}{R_{eq}} = \dfrac{6}{12}$

$\dashrightarrow \sf \dfrac{1}{R_{eq}} = \dfrac{1}{2}$

$\dashrightarrow \sf R_{eq} = 2 \: ohms$

» At constant temperature the current flowing through a conductor is directly proportional to the potential difference across its ends.

Formula : V = RI

where,

• V denotes potential difference
• R denotes resistance
• I denotes current

substituting all the given values in the formula,

$\dashrightarrow \sf V = RI$

$\dashrightarrow \sf 6 = (2)I$

$\dashrightarrow \sf I = \dfrac{6}{2}$

$\dashrightarrow \sf I = 3A$

Thus, the current flowing through the circuit is 3A.