Question #86804

1 Answer
Feb 27, 2017

Answer:

(a) #R_1=200Omega, R_2=160Omega, R_3=80Omega, and R_4=200Omega#
(b) #I=110" mA"#

Explanation:

(a) We know that effective resistance of two resistors connected in series is sum of individual resistances
#R_s=R_1+R_2#
and for two resistors connected in parallel is given by the expression
#R_p=(R_1R_2)/(R_1+R_2)#

Keeping above in mind the effective resistance of the circuit is
#R_e=R_1+((R_2+R_3)R_4)/((R_2+R_3)+R_4)# ......(1)
Given is #R_2 = 2R_3, R_1 = R_4, 2R_1 = 5R_3#

Inserting in (1) above and making all in terms of #R_3 and R_1#
#R_e=R_1+((2R_3+R_3)R_1)/((2R_3+R_3)+R_1)#
#=>R_e=R_1+(3R_3R_1)/(3R_3+R_1)#
#=>R_e=R_1+(3(2/5R_1)R_1)/(3(2/5R_1)+R_1)#
#=>R_e=R_1+(6/5R_1^2)/(11/5R_1)#
#=>R_e=R_1+6/11R_1#
#=>R_e=17/11R_1#

Assuming that internal resistance of the voltage source is small and is ignored in the calculations
#I=V/R#
#=>I_1=34/R_e#
#=>0.11=34/(17/11R_1)#
#=>R_1=(34xx11)/(17xx0.11)#
#=>R_1=200Omega#
Now #R_3=2/5R_1#
#=>R_3=2/5xx200=80Omega#
Also #R_2=2R_3#
#R_2=2xx80=160Omega#
And #R_1=R_4=200Omega#

(b) As seen from the circuit, both the Resistor #R_1# and the voltage source are in series. In series circuit current is same in all the circuit components.
#:.I=I_1=110" mA"#

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Text (11.9) [T/I] in the question is not relevant to the circuit.