Question

Two resistors, R 1 and R 2 , are connected in parallel. R 2 =221.0 ohms, and the equivalent resistance of the combination is 120.7 ohms. What is the value of R 1 ? (Unit = ohm)​

Answer 1

Answer:

$\rm { R_1 }$ = 265.9Ω

Explanation:

As per the provided information in the given question, we have :

• $\rm { R_2 }$ = 221.0Ω
• $\rm { R_P }$ = 120.7Ω
• $\rm { R_1 }$ and $\rm { R_2 }$ are connected in parallel combination.

We are asked to calculate the value of of $\rm { R_1 }$.

As we know that, when the resistors are connected in parallel combination, then equivalent resistance in the circuit is given by :

$\\ \twoheadrightarrow \quad \pmb{\boxed{\sf { \dfrac{1}{R_p} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dots \dfrac{1}{R_n} } }}$

Substituting values,

$\\ \twoheadrightarrow \quad \sf {\dfrac{1}{R_p} = \dfrac{1}{R_1} + \dfrac{1}{R_2} }$

$\\ \twoheadrightarrow \quad \sf {\dfrac{1}{120.7} = \dfrac{1}{R_1} + \dfrac{1}{221} }$

$\\ \twoheadrightarrow \quad \sf {\dfrac{1}{120.7} = \dfrac{221 + R_1}{(R_1)(221)} }$

$\\ \twoheadrightarrow \quad \sf {\dfrac{1}{120.7} = \dfrac{221 + R_1}{221R_1} }$

$\\ \twoheadrightarrow \quad \sf { 1(221R_1) = 120.7(221 + R_1)}$

$\\ \twoheadrightarrow \quad \sf { 221R_1 = 26674.7 + 120.7R_1 }$

$\\ \twoheadrightarrow \quad \sf { 221R_1 - 120.7R_1 = 26674.7 }$

$\\ \twoheadrightarrow \quad \sf { 100.3R_1 = 26674.7 }$

$\\ \twoheadrightarrow \quad \sf { R_1 = \cancel{\dfrac{26674.7}{100.3} }}$

$\\ \twoheadrightarrow \quad \bf \underline{ R_1 = 265.9 \; \Omega}$

Therefore, value of R_1 is 265.9Ω.

Points to remember :

When the resistors are connected in parallel combination, then equivalent resistance in the circuit is given by :

$\\ \twoheadrightarrow \quad \pmb{\boxed{\sf { \dfrac{1}{R_p} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dots \dfrac{1}{R_n} } }}$

When the resistors are connected in series combination, then equivalent resistance in the circuit is given by :

$\\ \twoheadrightarrow \quad \pmb{\boxed{\sf { R_S = R_1 + R_2 +\dots R_n} }}$