Question

Please solve guyz...
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Answer 1

refer to attachment..

(I think ,the book is supporting material..)

hope it's helpful to you.

Answer 2

Here, as we observed that three resistance is in parallel So,

Combined Resistance of three resistance connected in parallel.

$\big($$\large\mathtt{Quick\: \: Formula \: \: Sheet \: :} \big)$

In Parallel ,

• $\boxed{ \bf { \dfrac{1}{R_eq} = \dfrac{1}{R} + \dfrac{1}{R_3} }}$

Using Ohms law :

• $\dashrightarrow\:\:\sf V = IR$

Then,

• ${\boxed{\bf { I = \dfrac{V}{R} }}}$

Doing the answer of the above question now ;

Now, we have two resistance of 12 ohms each connected in parallel. This arrangement is shown in the attachment figure. We want to find out their equivalent resistance or sum resistance. We know that when two resistance considering them $\sf ( R_1 \: and \: R_2 )$ are connected in parallel, then their sum resistance R is given by :

Where,

$\sf R = Sum \: Resistance$

$\sf R_1 = 12 \: ohms$

$\sf R_2 = 12 \: ohms$

Putting these values , we get :

$\rightarrow \sf \dfrac{1}{R} = \dfrac{1}{12} + \dfrac{1}{12}$

$\rightarrow \sf \dfrac{1}{R} = \dfrac{1 + 1}{12}$

$\rightarrow \sf \dfrac{1}{R} = \dfrac{2}{12}$

$\rightarrow \sf \dfrac{1}{R} = \dfrac{1}{6}$

$\rightarrow \sf \dfrac{1}{R} = 6 \: Ohms$

Now, this one R is again connected in parallel with $\sf ( R_3 )$ then, their equivalent resistance is given by :

$\sf \dfrac{1}{R_eq} = \dfrac{1}{R} + \dfrac{1}{R_3}$

Here,

• $\sf R = 6 \: Ohms$

• $\sf {R_3 = 12 \: Ohms}$

$\rightarrow \sf \dfrac{1}{R_eq} = \dfrac{1}{6} + \dfrac{1}{12}$

$\rightarrow \sf \dfrac{1}{R_eq} = \dfrac{2 + 1}{12}$

$\rightarrow \sf \dfrac{1}{R_eq} = \dfrac{3}{12}$

$\rightarrow \sf \dfrac{1}{R_eq} = \dfrac{1}{4}$

$\rightarrow \sf \dfrac{1}{R_eq} = 4 \: Ohms$

For Finding out the current, we have to know that the current is nothing but voltage of the circuit divided by total resistance or resultant resistance.

That is ,

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

Where,

• $\longrightarrow\:\:\sf I = Current$
• $\longrightarrow\:\:\sf V = Voltage$
• $\longrightarrow\:\:\sf R = Equivalent \: Resistance$

Putting the values , we get :

$\dashrightarrow\:\:\sf I = \dfrac{3}{4}$

$\dashrightarrow\:\:\sf I = 0.75 \: A$

$\dashrightarrow\:\: \underline { \boxed {\bf{ Current = 0.75 \: Amperes}}}$

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$\underline { \sf {Data \: analysis}}$

⠀$\underline{\underline{\maltese\:\: \textbf{\textsf{Combination \: of \: resistance \: or(Resistor) }}}}$

The resistance can be combined in two ways :

→ In Series

→ In Parallel

⠀⠀Resistances in Series

The combined resistance of any number of resistances connected in series is equal to the sum of all the individual resistance.

Formula :

${\boxed{\bf{R = R_1 + R_2 ..}}}$

Points to keep in mind :

★ ➤ When resistance connected in series are joined to the terminals of battery, then each resistance has a different potential difference ( V ) .

★ ➤ When resistance connected in series are joined to the terminals of battery, then current in each resistance will remain same.

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⠀⠀⠀⠀ Resistance in Parallel

The reciprocal of the combined resistance of a number of resistances connected in parallel is equal to the sum of the reciprocals of all the individual resistances.

Formula :

${\boxed{\bf {\dfrac{1}{R_eq} = \dfrac{1}{R} + \dfrac{1}{R_3}.. }}}$

Points to keep in mind :

★ ➤ When resistance connected in parallel are , then potential difference ( V ) across each resistance is the same which is equal to the voltage of the battery applied.

★ ➤ When resistance connected in parallel are joined to the two terminals of battery, then diffrent amount of current flow through each resistance.

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