Question

$\huge{\underline{\underline{\mathfrak{Question}}}}$


An electronics hobbyist is building a radio which requires 150 Ω in her circuit, but she has only 220 Ω, 79 Ω and 92 Ω resistors available. How can she connect the available resistors to get desired value of resistance?​

Answer 1

Answer:

She can do so by connecting the 220 and 79 ohm resistors in parallel and connecting the 92 ohm resistor in parallel to the combination of the 220 ohm and 79 ohm resistors in parallel. (Refer to the attachment.)

Explanation:

Resistance between points B and  C = [1/220 + 1/79]⁻¹

= [(79 + 220)/220*79]⁻¹

= [399/17380]⁻¹

= 17380/399

= 58.12 Ω

Resistance between points A and C = 92 + 58.12 = 150.12Ω

Hence justified.

Answer 2

Answer:

Given :

Resistors available :

• 220 ohm
• 79 ohm
• 92 ohm

To find:

Combination of resistances such that the Equivalent resistance is approximately 150 ohms.

Concept:

First of all, the person has to connect the 220 ohm and 79 ohm in parallel connection. This should be connected in parallel with 92 ohms to get the required resistance of 150 ohm.

Calculation:

Net resistance in Parallel connection :

$\therefore \dfrac{1}{R} = \dfrac{1}{R1} + \dfrac{1}{R2}$

$\implies \dfrac{1}{R} = \dfrac{1}{220} + \dfrac{1}{79}$

$\implies \: R = \dfrac{(220 \times 79)}{220 + 79}$

$\implies \: R = \dfrac{17380}{299}$

$\implies \: R = 58.322 \: ohm$

Now this is connected in series with 92 ohms.

So , equivalent resistance be :

$\therefore \: r_{eq} \: = 58.322 + 92$

$\implies\: r_{eq} \: = 150.322 \: ohm$

$\implies\: r_{eq} \: \approx 150 \: ohm$

Diagram:

Please see attached the photo to understand better.