In the circuit shown in Fig., find the value of `R_`.

Explanation

Let the current through the various arms be as shown in Fig. Follow from the solution of question 39, `I_B//I_C=.01//1=1//100`
`I_(E)=I_(B)+I_(C)=I_C/100=I_C~~I_C`
In closed circuit `AB_(1)CEDA`
`I_(C)R_(C)+V_(CE)+I_(E)R_(E)=V_(C C) or I_(C)(R_C+R_(E))=V_(C C)-V_(CE)=12-3=9 ...(i)`
In closed circuit AB_(1)GHJDA
`(I_B+I)R_B+IR+IR=V_(C C)=12 or I_(B)I_(R)+I(R_(B)+R)=12`
or `I_C/betaR_(B)+I(R_(B)+R)=12 ...(ii)`
Let V be the potential difference across H and J, then
`V=IR=V_(BE)+I_(E)R_(E)=V_(BE)+I_(C)R_(E)=0.5+I_(C)R_(E) or I=(0.5+I_(C)R_(E)//R`
Putting this value of I in (ii), we get
`I_C/betaR_(B)+1/R(1/2+I_(C)R_(E))(R_(B)+R)=12 or (I_C R_B)/beta+1/R[1/2R_B+1/2R+I_CR_(E)R_B+I_CR_(E)R]=12`
or `I_C R_R + beta [1/2R_(B)+1/2R+I(C)_R_(E)R_(B)+I_(C)R_(E)R]=12beta R`
or `I_C[R_(B)R+beta R_(E)(R_(B)+R)]+1/2beta(R_(B)+R)=12betaR`
or `I_(C)[(100xx10^(3))(20xx10^(3))+100xx(10^(3))(100xx10^(3)+20xx10^(3))]+1/2xx100[100xx10^(3)+20xx10^(3)]=12xx100xx20xx10^(3)`
or `I_(C)[2xx10^(9)+12xx10^(9)]+6xx10^(6)=24xx10^(6) or I_(C)=((24-6)10^(6))/((12+2)xx10^(9))=9/7xx10^(-3)A`
Putting value of I_ in (i), we get
`9/7xx10^(-3)(R_(C)+10^(3))=9 or R_(C)+10^(3)=7xx10^(3) or R_(C)=7xx10^(3)-10^(3)=6xx10^(3)Omega=6kOmega`