The current passing through the $3 Ω$ resistor is [[1]] $A$ if both the keys of the circuit are closed.

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Detailed Solution

The current passing through the

3 Ω

resistor is

1.2 A

if both the keys of the circuit are closed.
If both the keys are closed then due to parallel combination the potential difference across the series connection of a resistor

3 Ω

and

2 Ω

6 V

.
Now, the equivalent resistance of series connection is

R_{S} = 2 + 3

{\Rightarrow R}_{S} = 5 Ω

From the Ohm’s law, the current flowing through the

3 Ω

resistor is,

\Rightarrow I = \frac{V}{R_{S}}

\Rightarrow I = \frac{6}{5}

\Rightarrow I = 1.2 A

The current flowing through the

3 Ω

resistor is

1.2 A

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