Two resistances R1 and R2 have effective resistance Rs when connected in series and Rp, when connected in parallel. If Rs Rp=16 and R1/R2=4, calculate the values of R1and R2 (in units of resistance).

Question

Two resistances R1 and R2 have effective resistance Rs when connected in series and Rp, when connected in parallel. If Rs Rp=16 and R1/R2=4, calculate the values of R1and R2 (in units of resistance). (A) R1=2,R2=0.5 (B) R1=1,R2=0.25 (C) R1=8,R2=2 (D) R1=4,R2=1

Tardigrade Admin · Accepted Answer

In series Rs = R1 + R2 In parallel Rp = (R1R2/R1 + R2) Here, R1 4 R2 (given) and RsRp = 16 (given) But Rs Rp = ( R1 + R2) (R1R2/R1 +R2) = R1R2 = 16 i.e. 4R2 × R2 = 16 i.e. R2 = 2 Ω and R1 = 8 Ω

Q. 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).

$R_{1} = 2, R_{2} = 0.5$

$R_{1} = 1, R_{2} = 0.25$

$R_{1} = 8, R_{2} = 2$

$R_{1} = 4, R_{2} = 1$

Q. Two resistances R1​ and R2​ have effective resistance Rs​ when connected in series and Rp​, when connected in parallel. If Rs​Rp​=16 and R1​/R2​=4, calculate the values of R1​and R2​ (in units of resistance).

R1​=2,R2​=0.5

R1​=1,R2​=0.25

R1​=8,R2​=2

R1​=4,R2​=1

In series Rs​=R1​+R2​ In parallel Rp​=R1​+R2​R1​R2​​ Here, R1​4R2​ (given) and Rs​Rp​ = 16 (given) But Rs​Rp​=(R1​+R2​)R1​+R2​R1​R2​​=R1​R2​ = 16 i.e. 4R2​×R2​ = 16 i.e. R2​=2Ω and R1​=8Ω

Q. 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).

$R_{1} = 2, R_{2} = 0.5$

$R_{1} = 1, R_{2} = 0.25$

$R_{1} = 8, R_{2} = 2$

$R_{1} = 4, R_{2} = 1$