Answer to Question #155914 in Electric Circuits for Jerick

Question

Three resistances of 12, 16, and 20 ohms are connected in parallel. What resistance must be connected in series with this combination to give a total resistance of 25 ohms?

Answer 1

Let's first find the total resistance of three resistances connected in parallel:

$\dfrac{1}{R_{tot}}=\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}}+\dfrac{1}{R_{3}},$ $R_{tot}=\dfrac{1}{\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}}+\dfrac{1}{R_{3}}},$ $R_{tot}=\dfrac{1}{\dfrac{1}{12\ \Omega}+\dfrac{1}{16\ \Omega}+\dfrac{1}{20\ \Omega}}=5\ \Omega.$

Finally, we can find the resistance that must be connected in series with this combination to give a total resistance of 25 ohms:

$R_{tot,parallel}+R=R_{tot,series},$ $R=R_{tot,series}-R_{tot,parallel}=25\ \Omega-5\ \Omega=20\ \Omega.$

Answer:

$R=20\ \Omega.$