Question

please give this answer correctly ​

Answer 1

Given :-

• R1 = R2 = R3 = 30Ω

• Voltage, v = 2 v

To find :-

• Value of I, current in the circuit

Solution :-

In the given figure, we can see that resistors R1 and R2 are in series connection.

Their equivalent resistance will be:

$: \implies \sf {R_{(s)} =R1 +R2 }$

${ : \implies \sf {R_{(s)} = 30 \Omega + 30\Omega}}$

${ : \implies \sf {R_{(s)} = 60 \Omega }}$

Now this equivalent connection of R1 and R2 is in parallel connection with R3.

${ : \implies\sf\dfrac{1}{R_{(equivalent)}}=\dfrac{1}{R_{(s)}}+\dfrac{1}{R_3}}$

${ : \implies \sf\dfrac{1}{R_{(equivalent)}}=\dfrac{1}{60\Omega}+\dfrac{1}{30\Omega}}$

${ : \implies\sf\dfrac{1}{R_{(equivalent)}}=\dfrac{1 + 2}{60\Omega}}$

${ : \implies\sf\dfrac{1}{R_{(equivalent)}}=\dfrac{3}{60\Omega}}$

${ : \implies\sf\dfrac{1}{R_{(equivalent)}}= \cancel\dfrac{3}{60\Omega}}$

${ \implies \sf\dfrac{1}{R_{(equivalent)}}=\dfrac{1}{20\Omega}}$

$\sf : \implies R_{(equivalent)}=20\Omega$

So the equivalent resistance of whole circuit is 20Ω.

Now applying ohm's law :-

$\leadsto\sf V = IR$

Substitute values

$\sf \leadsto 2v = I(20\Omega)$

$\sf \leadsto \dfrac{ 2v} {20\Omega} =I$

$\sf \leadsto \cancel \dfrac {2v} {20\Omega} =I$

$\sf \leadsto \dfrac {1v} {10\Omega} =I$

$\sf \leadsto 0.1 \: Ampere = I$

So the current I flowing through the circuit is 0.1 Ampere.

[Omega will change into Ω when viewed from website]

Extra :-

• Resistors in series : Resistors are said to be in series connection when more than 1 resistors are connected from end to end.

• To find resistance in series connection, we apply formula , R = R1 + R2 + R3 . . .

• Resistors in parallel : Resistors are said to be in a parallel connection when the end points of two or more resistors are connected to a same point.

• To find resistance in parallel connection, we apply formula 1/R = 1/R1 + 1/R2 + 1/R3 . . .

• Current : Remains same in series connection of resistors.

• Voltage : Remains same in parallel connection of resistors.