Question

*Find the value of equivalent resistance if three resistance of15 ohm, 20ohm,10 ohm are connected in parallel combination.*

1️⃣ 46 ohm
2️⃣ 0.46 ohm
3️⃣ 0.046 ohm
4️⃣ 4.6 ohm​

Answer 1

Let

R_1 = 15 ohm

R_2 = 20 ohm

R_3 = 10 ohm

When the resistors are connected in parallel, the net resistance is given by,

$\dfrac{1}{R} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3} \\ \\ \dfrac{1}{R} = \dfrac{1}{15} + \dfrac{1}{20} + \dfrac{1}{10} \\ \\ \dfrac{1}{R} = \dfrac{1}{15} + \dfrac{1}{20} + \dfrac{(1 \times 2)}{(10 \times 2)} \\ \\ = \dfrac{1}{15} + \dfrac{1}{20} + \dfrac{2}{20} \\ \\ = \dfrac{1}{15} + \dfrac{(1 + 2)}{20} \\ \\ = \dfrac{1}{15} + \dfrac{3}{20} \\ \\ = \dfrac{20 + (3 \times 15)}{20 \times 15} \\ \\ = \dfrac{20 + 45}{20 \times 15} \\ \\ \dfrac{1}{R} = \dfrac{65}{20 \times 15} \\ \\ R = \dfrac{20 \times 15}{65} \\ \\ R = \dfrac{300}{65} \\ \\ R = 4.62 \: ohm$

The net resistance is 4.6 ohm.

Option 4 is right.

Answer 2

Required Answer:

→ Option 4

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◆ Procedure to solve this:

Here, in this question we are given that three resistance are connected in parallel combination.This means, here we have to apply the formula given below:

• ${\underline {\boxed {\large {\bf \gray { \dfrac{1}{R_p} = \dfrac{1}{R_1} + \dfrac{1}{R_2}.......\dfrac{1}{R_n} } }}}}$

● Equivalent resistance in the parallel is equal to the reciprocal of the sum of reciprocal of the resistances.

So, by putting the values in the above formula, we can find the equivalent resistance.

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Let's solve it!

Given:

• Resistances are in parallel combination.
• $\sf { R_1}$ = 15 Ω
• $\sf { R_2}$ = 20 Ω
• $\sf { R_3}$ = 10 Ω

To find:

• Equivalent resistance ( R )

Calculation:

We know that,

• $\sf { \dfrac{1}{R_p} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3} }$

Substituting values:

$\sf { \longrightarrow \dfrac{1}{R_p} = \dfrac{1}{15} + \dfrac{1}{20} + \dfrac{1}{10} }$

$\sf { \longrightarrow \dfrac{1}{R_p} = \dfrac{4+3+6}{60}}$

$\sf { \longrightarrow \dfrac{1}{R_p} = \dfrac{13}{60}}$

$\sf { \longrightarrow R_p = \dfrac{60}{13}}$

$\sf { \longrightarrow R_p = 4.61 \: ohm }$

Thus, Option 4 is correct.

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More :

● When resistances are connected in parallel combination:

• ${\underline {\boxed {\large {\bf \gray { \dfrac{1}{R_p} = \dfrac{1}{R_1} + \dfrac{1}{R_2}.......\dfrac{1}{R_n} } }}}}$

● When resistances are connected in series combination:

• ${\underline {\boxed {\large {\bf \gray { R_s = R_1 + R_2 +......R_n } }}}}$

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