Question

Find the equivalent resistance for 20ohm 15ohm and 20ohm parallel conecetion​

Answer 1

$\huge\mathtt{\fbox{\red{Answer✍︎}}}$

GIVEN :-

Three resistors of 20Ω , 15Ω , 20Ω.

TO FIND :-

The equivalent resistance.

SOLUTION :-

Let R₁ be 20Ω R₂ be 15Ω R₃ be 20Ω.

Now as we know that , when the resistors are connected in parallel combination then their equivalent resistance is given by,

$\\ : \implies \displaystyle \sf \: \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + ... + \frac{1}{R_n}$

$: \implies \displaystyle \sf \: \frac{1}{R_{eq}} = \frac{1}{20} + \frac{1}{15} + \frac{1}{20}$

$: \implies \displaystyle \sf \: \frac{1}{R_{eq}} = \frac{3 + 4 + 3}{60}$

$: \implies \displaystyle \sf \: \frac{1}{R_{eq}} = \frac{10}{60}$

$: \implies \displaystyle \sf \: \frac{1}{R_{eq}} = \frac{1}{6}$

$: \implies \displaystyle \underline{ \boxed{\sf \bold{ \: {R_{eq}} = 6 \: \Omega}}}$

Answer 2

Answer:

GIVEN :-

Three resistors of 20Ω , 15Ω , 20Ω.

TO FIND :-

The equivalent resistance.

SOLUTION :-

Let R₁ be 20Ω R₂ be 15Ω R₃ be 20Ω.

Now as we know that , when the resistors are connected in parallel combination then their equivalent resistance is given by,

$\begin{gathered} \\ : \implies \displaystyle \sf \: \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + ... + \frac{1}{R_n} \\ \\ \\ \end{gathered} </p><p>:⟹ </p><p>R </p><p>eq</p><p> </p><p> </p><p>1</p><p> </p><p> = </p><p>R </p><p>1</p><p> </p><p> </p><p>1</p><p> </p><p> + </p><p>R </p><p>2</p><p> </p><p> </p><p>1</p><p> </p><p> + </p><p>R </p><p>3</p><p> </p><p> </p><p>1</p><p> </p><p> +...+ </p><p>R </p><p>n</p><p> </p><p> </p><p>1</p><p> </p><p> </p><p> </p><p> </p><p></p><p>\begin{gathered} : \implies \displaystyle \sf \: \frac{1}{R_{eq}} = \frac{1}{20} + \frac{1}{15} + \frac{1}{20} \\ \\ \\ \end{gathered} </p><p>:⟹ </p><p>R </p><p>eq</p><p> </p><p> </p><p>1</p><p> </p><p> = </p><p>20</p><p>1</p><p> </p><p> + </p><p>15</p><p>1</p><p> </p><p> + </p><p>20</p><p>1</p><p> </p><p> </p><p> </p><p> </p><p></p><p>\begin{gathered}: \implies \displaystyle \sf \: \frac{1}{R_{eq}} = \frac{3 + 4 + 3}{60} \\ \\ \\ \end{gathered} </p><p>:⟹ </p><p>R </p><p>eq</p><p> </p><p> </p><p>1</p><p> </p><p> = </p><p>60</p><p>3+4+3</p><p> </p><p> </p><p> </p><p> </p><p></p><p>\begin{gathered}: \implies \displaystyle \sf \: \frac{1}{R_{eq}} = \frac{10}{60} \\ \\ \\ \end{gathered} </p><p>:⟹ </p><p>R </p><p>eq</p><p> </p><p> </p><p>1</p><p> </p><p> = </p><p>60</p><p>10</p><p> </p><p> </p><p> </p><p> </p><p></p><p>\begin{gathered}: \implies \displaystyle \sf \: \frac{1}{R_{eq}} = \frac{1}{6} \\ \\ \\ \end{gathered} </p><p>:⟹ </p><p>R </p><p>eq</p><p> </p><p> </p><p>1</p><p> </p><p> = </p><p>6</p><p>1</p><p> </p><p> </p><p> </p><p> </p><p></p><p>: \implies \displaystyle \underline{ \boxed{\sf \bold{ \: {R_{eq}} = 6 \: \text{\O}mega}}}:⟹ </p><p>R </p><p>eq</p><p> </p><p> =6Ømega</p><p> </p><p> </p><p>$