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What will be the ratio of the power used in the 2 Ω resistor in each of the following circuits, as shown in the figure below?

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1 : 1

1 : 2

2 : 1

1 : 4

answer is A.

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Detailed Solution

The ratio of the power used in the

2 Ω

resistor in each of the following circuits is

1 : 1

Case (i) -

Voltage (V) = 6 V

R_{1} = 1 Ω

R_{2} = 2 Ω

As in figure (a)

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 Ω + 2 Ω

R_{s} = 3 Ω

We know that, Current

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

…………. (ii)
Now on putting values in equation (ii) we get

I = \frac{6 V}{3 Ω}

I = 2 A

Now power across

2 Ω

is;

Power (P) = I^{2} R

= 2^{2} \times 2

P = 2^{2} \times 2

P = 4 \times 2 W

P = 8 W

…………………………. (iii)
Case (ii)

Voltage (V) = 4 V

R_{1} = 12 Ω

R_{2} = 2 Ω

As in figure (b)

R_{1}

and

R_{2}

both are connected in parallel, so the potential difference of each resistor remains the same. So, the power across

2 Ω

P = \frac{V^{2}}{R}

= > P = \frac{4^{2}}{2}

= > P = \frac{4 \times 4}{2}

= > P = 8 W

…………………………….. (iv)
So, the power across the parallel combination is

8 W .

On dividing equation (iii) by (iv) we get the ratio of power in series to power in parallel is

= \frac{8}{8}

= 1 : 1

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