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The current which is flowing through the resistor $3 Ω$ is equal to [[1]] $A$ .

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Detailed Solution

The current which is flowing through the resistor

3 Ω

is equal to

0.67 A

.
It is given that the

3 Ω

and

6 Ω

are connected in parallel which are in turn connected in series with a

10 Ω

and the

12 V

power supply.
Since, resistors of resistances

3 Ω

and

6 Ω

are connected in parallel, the reciprocal of the equivalent resistance is equal to the sum of the reciprocals of the individual resistances. Let the equivalent resistance of

3 Ω

and

6 Ω

R_{P}

. Therefore, the equivalent resistance of

3 Ω

and

6 Ω

is calculated as

\frac{1}{R_{P}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}

\Rightarrow \frac{1}{R_{P}} = \frac{1}{3} + \frac{1}{6}

\Rightarrow \frac{1}{R_{P}} = \frac{1}{2}

\Rightarrow R_{P} = 2 Ω

Also, the resistance

R_{P}

and

10 Ω

are connected in series therefore their equivalent resistance is the sum of the individual resistance. The equivalent resistance is given by

R = R_{P} + 10

\Rightarrow R = 2 + 10

\Rightarrow R = 12 Ω

Therefore, from the Ohm’s law, the total current in the circuit is

I = \frac{V}{R}

\Rightarrow I = \frac{12}{12}

\Rightarrow I = 1 A

The potential difference across the resistors

3 Ω

and

6 Ω

V_{P} = I R_{P}

\Rightarrow V_{P} = 1 \times 2

\Rightarrow V_{P} = 2 V

The current flowing through the resistor

3 Ω

is,

I = \frac{V_{P}}{R}

\Rightarrow I = \frac{2}{3}

\Rightarrow I = 0.67 A

Hence, the current flowing through the resistor

3 Ω

0.67 A

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