Question

Two resistances $R_1$ and $R_2$ have effective resistance $R_s$ when connected in series and $R_p$, when connected in parallel. If $R_s R_p=16$ and $R_1/R_2=4$, calculate the values of $R_1$and $R_2$ (in units of resistance).

• A. $R_1=2,R_2=0.5$
• B. $R_1=1,R_2=0.25$
• C. $R_1=8,R_2=2$
• D. $R_1=4,R_2=1$

Answer 1

Answer: (C)

$R_1=8,R_2=2$

Answer 2

In series $R_s = R_1 + R_2$ In parallel $R_p = \frac{R_1R_2}{R_1 + R_2}$ Here, $R_1 4 \, R_2$ (given) and $R_sR_p$ = 16 (given) But $R_s R_p = ( R_1 + R_2) \frac{R_1R_2}{R_1 +R_2} = R_1R_2$ = 16 $i.e.$ $4R_2 \times R_2$ = 16 $i.e.$ $R_2 = 2 \, \Omega$ and $R_1 = 8 \, \Omega$