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The parallel combination of two $10 Ω$ resistors are placed across the terminals of a $24 V$ battery, the current through each branch in the circuit is [[1]] $A$ and [[2]] $A$ .

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Detailed Solution

When the parallel combination of two

10 Ω

resistors are placed across the terminals of a

24 V

battery, the current through the circuit is

4.8 A

.
It is given that two resistors of equal resistance namely,

R_{1} = 10 Ω

R_{2} = 10 Ω

are placed across the battery of potential difference,

V = 24 V

From the Ohm’s law, the current

(I)

flowing through the circuit is

I = \frac{V}{R}

Where

V

and

R

are the potential difference and the resistance respectively.
Now, the current through the resistance of

10 Ω

I_{1} = \frac{V}{R_{1}}

\Rightarrow I_{1} = \frac{24}{10}

\Rightarrow I_{1} = 2.4 A

Hence, the current through the resistance

10 Ω

2.4 A

.
Similarly, the current through the resistance of

10 Ω

I_{2} = \frac{V}{R_{2}}

\Rightarrow I_{2} = \frac{24}{10}

\Rightarrow I_{2} = 2.4 A

Hence, the current through the resistance

10 Ω

2.4 A

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