Answer:
Explanation:
When resistors are connected in parallel combination, then equivalent resistance is given by,
[tex] \to \quad \bf{ \dfrac{1}{R_1} + \dfrac{1}{R_2} =\dfrac{1}{R_p}}[/tex]
Inserting values,
[tex] \to \quad \bf { \dfrac{1}{R_1} + \dfrac{1}{R_2} =\dfrac{1}{R_p} } \\ \\ \to \quad \bf { \dfrac{1}{R_1} + \dfrac{1}{221} =\dfrac{1}{120.7} } \\ \\ \to \quad \bf { \dfrac{221+R_1}{(R_1)(221)} =\dfrac{1}{120.7} } \\ \\ \to \quad \bf { \dfrac{221+R_1}{221R_1} =\dfrac{1}{120.7} } \\ \\ \to \quad \bf { 120.7(221+R_1) = 221R_1(1)} \\ \\ \to \quad \bf { 26674.7 + 120.7R_1= 221R_1} \\ \\ \to \quad \bf { 26674.7= 221R_1- 120.7R_1} \\ \\ \to \quad \bf { 26674.7 = 100.3R_1} \\ \\ \to \quad \bf { \dfrac{26674.7 }{100.3} =R_1} \\ \\ \to \quad\underline{\boxed{ \bf { 265.9 \; \Omega=R_1}}} \\ [/tex]
