Question

Draw a schematic Diagram of a circuit consisting of a battery of three cells of 2v
each a 5Ω an 8Ω resistor and a 12 Ω resistor and plug key all connected in series
also find current flow through 12Ω resistor.​

Answer 1

diagram of a circuit consisting of a battery of three cells of 2v

Answer 2

$\displaystyle\large\underline{\sf\red{Given}}$

✭ A circuit consists of,

• Three 2V batteries
• A 5Ω, 8Ω & 12Ω resistors are connected in series

$\displaystyle\large\underline{\sf\blue{To \ Find}}$

◈ Current through the 12Ω resistor?

$\displaystyle\large\underline{\sf\gray{Solution}}$

Also in a series connection the potential difference will be

$\displaystyle\underline{\boxed{\sf Pd = V_1+V_2+V_3....+V_n}}$

And the Resistance will be given by,

$\displaystyle\underline{\boxed{\sf \dfrac{1}{R_{eq}} = \dfrac{1}{R_1}+\dfrac{1}{R_2}...\dfrac{1}{R_n}}}$

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$\underline{\bigstar\:\textsf{According to the given Question :}}$

So now we shall find the Pd across the circuit,

➝ $\displaystyle\sf Pd = V_1+V_2+V_3....+V_n$

➝ $\displaystyle\sf Pd = 2+2+2$

➝ $\displaystyle\sf \orange{Pd = 6 \ V}$

So now the Resistance will be,

»» $\displaystyle\sf \dfrac{1}{R_{eq}} = \dfrac{1}{R_1}+\dfrac{1}{R_2}...\dfrac{1}{R_n}$

»» $\displaystyle\sf \dfrac{1}{R_{eq}} = \dfrac{1}{5}+\dfrac{1}{8}+\dfrac{1}{12}$

»» $\displaystyle\sf \dfrac{1}{R_{eq}} = \dfrac{1}{5}+\dfrac{1}{8}+\dfrac{1}{12}$

»» $\displaystyle\sf\purple{R_{eq} = 25\Omega}$

$\displaystyle\underline{\boxed{\sf V = IR}}$

• V = Potential difference (Pd) = 6 V

• I = Current = ?

• R = Resistance = 25Ω

Substituting the values,

➳ $\displaystyle\sf V = IR$

➳ $\displaystyle\sf 6 = I \times 25$

➳ $\displaystyle\sf 6 = 25I$

➳ $\displaystyle\sf \dfrac{6}{25} = I$

➳ $\displaystyle\sf \pink{Current = 0.24 \ A}$

$\displaystyle\sf \therefore\:\underline{\sf Current \ accross \ the \ 12\Omega \ resistor \ is \ 0.24 A}$

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