Question

A flashlight uses two 1.5 V batteries to provide a steady current of 0.5 A in the filament. Determine the resistance of the glowing filament.
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Answer 1

$\Large{\bf{\green{\mathfrak{\dag{\underline{\underline{Given:-}}}}}}}$

• Flashlight uses two 1.5 V batteries to provide a steady current of 0.5 A in the filament.

$\Large{\bf{\orange{\mathfrak{\dag{\underline{\underline{To \: Find:-}}}}}}}$

• Resistance of the glowing filament.

$\Large{\bf{\red{\mathfrak{\dag{\underline{\underline{Solution:-}}}}}}}$

First,

• The flashlight uses two 1.5 V batteries to provide steady current.

Here,

• Total Voltage = 1.5 * 2 (Since, 2 batteries)

• Total Voltage = 3 V.

• V = 3V

Now,

• Current provided to the filament = 0.5 A

Therefore,

• I = 0.5A.

Now,

• Resistance offered by the glowing filament =

We know that,

$\boxed{\pink{\sf I \: = \: \dfrac{V}{R}}}$

Rearranging,

$\boxed{\pink{\sf R \: = \: \dfrac{V}{I}}}$

• Here,

• I = 0.5A

• V = 3V

Substituting the values,

• $\sf R \: = \: \dfrac{3}{0.5}$

• $\sf R \: = \: \dfrac{3}{\dfrac{5}{10}}$

• $\sf R \: = \: 3 \: * \: \dfrac{10}{5}$

• $\sf R \: = \: 3 \: * \: 2$

• $\sf R \: = \: 6 Ω$

Therefore,

• Resistance offered by the glowing filament = 6 Ω.

$\Large{\bf{\purple{\mathfrak{\dag{\underline{\underline{Extra \: Information:-}}}}}}}$

$\sf I \: = \: \dfrac{V}{R}$

• where,

• I is Current

• V is Potential difference

• R is Resistance

Answer 2

Answer:

Given :-

A flashlight uses two 1.5 V batteries to provide a steady current of 0.5 A in the filament

To Find :-

Resistance

Solution :-

We know that

V = IR

V = Potential difference

I = Current

R = Resistance

The total potential difference consumed = 1.5 × 2 = 3 V

Now

V = IR

3 = 0.5(R)

3/0.5 = R

6 = R

Hence,

The resistance is 6 Ω