Question

Question #86804

Answer 1

(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.