# Question Solved1 Answerquestion 1 &2 pls In this lab, you will have to predict and measure the equivalent resistance between the positions indicated by A, B, C and D in the diagram below. R B R2 АО connoo w R3 1. Redraw the circuit in order to be able to find an expression for the equivalent resistance in each case. 2. Find the expressions for equivalent resistance. Between A and B Between A and C Between A and D Drawing Expression Between Band C Between B and D Between C and D Drawing Expression

question 1 &2 pls

Transcribed Image Text: In this lab, you will have to predict and measure the equivalent resistance between the positions indicated by A, B, C and D in the diagram below. R B R2 АО connoo w R3 1. Redraw the circuit in order to be able to find an expression for the equivalent resistance in each case. 2. Find the expressions for equivalent resistance. Between A and B Between A and C Between A and D Drawing Expression Between Band C Between B and D Between C and D Drawing Expression
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Transcribed Image Text: In this lab, you will have to predict and measure the equivalent resistance between the positions indicated by A, B, C and D in the diagram below. R B R2 АО connoo w R3 1. Redraw the circuit in order to be able to find an expression for the equivalent resistance in each case. 2. Find the expressions for equivalent resistance. Between A and B Between A and C Between A and D Drawing Expression Between Band C Between B and D Between C and D Drawing Expression
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given:equivalent resistance(i) Between A and Bin that case R_(4) are not use in this circuit because R_(4) are not make a close circuit so no any curren flow in this resistancerarrR_(2)&amp;R_(3) resistance are in series so total resistance of R_(2)&amp;R_(3) is=> now R_(1) are parellel to R_(2)+R_(3) then (1)/(R_(eq))=(1)/(R_(1))+(1)/(R_(2)+R_(3)){:[(1)/(R_("eq "))=(R_(2)+R_(3)+R_(1))/(R_(1)(R_(2)+R_(3)))],[R_("eq ")=(R_(1)(R_(2)+R_(3)))/(R_(2)+R_(3)+R_(1))]:}(ii) Between A andR_(4) celso not considerrarrR_(1)&amp;R_(2) are in series combination and R_(3) are parellel to R_(1)+R_(2){:[" Q "(1)/(R_(eq))=(1)/(R_(1)+R_(2))+(1)/(R_(3))=(R_(3)+R_(1)+R_(2))/(R_(3)(R_(1)+R_(2)))],[R_(eq)=(R_(3)(R_(1)+R_(2)))/(R_(1)+R_(2)+R_(3))quad AA_(R_(1)+R_(2))ubrace(0)_(R_(3)^(0))]:}(iii) Between A and DR_(1) &amp; R_(2) are in series. so total Resista =R_(1)+R_(2)R_(1)+R_(2)+R_(3) are a parellel so to tal Resistanea{:[(1)/(r)=(1)/(R_(1)+R_(2))+(1)/(R_(3))=(R_(3)+R_(1)+R_(2))/(R_(3)(R_(1)+R_(2)))],[r=(R_(3)(R_(1)+R_(2)))/(R_(1)+ ... See the full answer