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What will be the power across $2$ resistors when a $6 V$ battery is connected in series with $1 Ω$ and $2 Ω$ resistors, as shown in the figure?

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8 W

10 W

12 W

16 W

answer is A.

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Detailed Solution

The power across

2 Ω

resistors when a

6 V

battery is connected in series with

1 Ω

and

2 Ω

resistors is

8 W .

Given,

Voltage (V) = 6 V

R_{1} = 1 Ω

R_{2} = 2 Ω

R_{1}

and

R_{2}

both are connected in series then net resistance (

R_{s}

) of the circuit can be calculated by the formula,

R_{s} = R_{1} + R_{2}

………………. (i)

R_{s} = 1 + 2

R_{s} = 3 Ω

…………………… (ii)
Ohm's law states that,

V = IR

I = \frac{V}{R_{s}}

………………….. (iii)
Where, I = Current flowing through the circuit. Put the values of V and

R_{s}

in equation (iii) we get,

I = \frac{6}{3}

I = 2 A

Because there is no current division in series circuits, this current will travel through each circuit component. As a result, the current passing through the 2

Ω

is 2 A.
We can calculate power by using formula,

P = I^{2} R

……………. (iv)

P = {(2)}^{2} \times 2

P = 4 \times 2

P = 8 W

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