Question

Calculate the equivalent resistance when two resistances 3 ohm & 6 ohm are
connected in (a) Series (b) parallel
​

Answer 1

Answer:

a)9ohm (b) 2 ohm

Explanation:

a)In series it is directly added,

therefore 3+6=9ohm(ans)

b)In parallel its reciprocal is added,

therefore (1/3)+(1/6)

1/R =(2+1)/6

1/R =3/6

1/R =1/2

:.R=2ohm(ans)

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Answer 2

Answer:-

$R_s = 9\Omega$

$R_p = 2\Omega$

Given:-

$R_1 = 3\Omega$

$R_2 = 6 \Omega$

To find :-

The equivalent resistance when they connected in series and parallel.

Solution:-

In series combination the equivalent resistance is given by :-

$\boxed{\sf{R_s = R_1 + R_2 .....R_n}}$

Put the value,

$R_s = 3 \Omega + 6 \Omega$

$R_s = 9 \Omega$

Now, the equivalent resistance in parallel is given by :-

$\boxed{\sf{\dfrac{1}{R_p}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_n}}}$

Now,

$\implies$ $\dfrac{1}{R_p}=\dfrac{1}{3}+\dfrac{1}{6}$

$\implies$ $\dfrac{1}{R_p}= \dfrac{2+1}{6}$

$\implies$ $\dfrac{1}{R_p}= \dfrac{3}{6}$

$\implies$ $\dfrac{1}{R_p}=\dfrac {1}{2}$

$\implies$ $R_p = 2 \Omega$

hence,the equivalent resistance is series combination and parallel combination will be $9\Omega , 2\Omega$