In an antique automobile, a 6-V battery supplies a total of 48 W to two identical headlights in parallel. The resistance of each bulb is:

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05-07-2022
Physics

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In an antique automobile, a 6-V battery supplies a total of 48 W to two identical headlights in parallel. The resistance of each bulb is:

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Dipanshushukla55idn Dipanshushukla55idn · Answer 1 · 2022-07-15T04:07:47-04:00

Answer:

The resistance of the each bulb is 1.75 ohm.

Explanation:

The Power P supplied by the battery is given by the formula,

P=VI

where V is the voltage and I is the current.

Then the total current through given parallel circuit is I=P/V

Given P=48 W and V=6 V the value of I is

I=48 W/6 V

I= 8 A

Since both bulbs have same resistance and are connected in parallel so the current through each circuit will be half of the total current. Hence the current through each circuit I₀ is 4 A.

The Resistance R is calculated using formula R=V/I₀.

So the resistance is

R=6 V/4 A

R=1.75 ohm

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Rahulraj37idn Rahulraj37idn · Answer 2 · 2022-07-18T01:28:25-04:00

Answer:

The resistance of the bulb is 1.5 ohm.

Explanation:

The power P delivered to the whole circuit is given by the formula,

P=VI

Given P= 48 W and V= 6V, the current in the circuit is

I=P/V

I=48/6= 8 A.

Since the bulbs are identical and are in parallel, therefore the current through each parallel circuit is half of the total current. Therefore current I1 through each bulb is 4 A.

The resistance is given by the formula,

R=V/I1

Hence R=6/4 = 1.5 ohm

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