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Study the circuit shown in which three identical bulbs B1, B2 and B3 are connected in parallel with a battery of $4.5 V$ . Each bulb has a wattage of $1.5 W$ . The total resistance of the circuit is

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4.5 ohm

5.5 ohm

6.5 ohm

None

answer is A.

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Detailed Solution

Study the circuit shown in which three identical bulbs B1, B2 and B3 are connected in parallel with a battery of

4.5 V

. The total resistance of the circuit is

4.5 ohm .

It is given that each bulb has a wattage of

1.5 W

. But we know the relation between power, voltage and resistance as

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

.
Given that, Voltage

V = 4.5 V

, Power

P = 1.5 W

By substituting voltage and power, we get resistance in the circuit. It is given by the formula,

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

⇨

\frac{{(4.5)}^{2}}{1.5}

⇨

13.5 ohm

To find out current in the circuit, we know the relation,

Power P = IV

Substituting the values of V and R in the above equation, we get Current

I = 1 A

.
As the bulbs B1, B2 and B3 are connected in parallel, the voltage across all the components remains the same but the current is distributed equally among the three bulbs. So,

I_{1} = \frac{1}{3} A

We know that,

P = I^{2} R

⇨

1.5 = I^{2} R

⇨

R = \frac{1.5}{{(\frac{1}{3})}^{2}}

⇨

R = 13.5 ohm

In a parallel circuit, current remains the same, potential difference and resistance are distributed.
So,

R_{1} = R_{2} = R_{3} = R / 3

Total resistance in the circuit is

\frac{13.5}{3} = 4.5 ohm .

Thus, total resistance in the circuit is

4.5 ohm .

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