Question

When two resistors of resistances R1 and R2 are connected in parallel, the net resistance is 3ohm. When connected in series, its value is 16ohm. Calculate the values of R1 and R2.

Answer 1

Answer:

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Answer 2

Answer:

Hence, $R_1=12 \Omega$ and $R_2=4 \Omega$.

Explanation:

Step 1: When $R_1$ and $R_2$ are connected in parallel, net resistance $\left(R_p\right)$ is given by,

$\begin{aligned}& \frac{1}{R_p}=\frac{1}{R_1}+\frac{1}{R_2} \\& \frac{1}{R_p}=\frac{R_1 R_2}{R_1+R_2} \\& R_p=\frac{R_1 R_2}{R_1+R_9}=3 \Omega\end{aligned}$

Step 2: When $R_1$ and $R_2$ are connected in series, the net resistance $\left(R_s\right.$ )is given by,

$R_s=R_1+R_2=16 \Omega$

$\begin{aligned}& \frac{R_1 R_2}{R_1+R_2}=3 \\& \frac{R_1 R_2}{16}=3 \\& \left(\text { Since, } R_1+R_2=16 \Omega\right) \\& R_1 R_2=48 \\& \text { or } R_1\left(16-R_1\right)=48 \\& 16 R_1-R_1^2=48 \\& R_1^2-16 R_1+48=0 \\& \text { which gives, ( } \left.R_1-12\right)\left(R_1-4\right)=0 \\& \text { Hence, } R_1=12 \Omega \text { and } R_2=4 \Omega\end{aligned}$

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