Given the circuit below. R1 is 94 ohms, R2 is 114 ohms, R3 is 245 ohms, R4 is 14

Question

Given the circuit below. R1 is 94 ohms, R2 is 114 ohms, R3 is 245 ohms, R4 is 145 ohms, R5 is 213 ohms, R6 is 260 ohms, R7 is 293 ohms, and R8 is 207 ohms. The potential difference between points a and b is 26 V.

A) What is the potential difference across each resistor? Put in the value for R1 followed by a comma, the value for R2 followed by a comma, the value for R3 followed by a comma, ... and the value for R8.

B)What is the current through each resistor?Put in the value for R1 followed by a comma, the value for R2 followed by a comma, the value for R3 followed by a comma, ... and the value for R8.

Explanation / Answer

To calculate potential difference across each element u need to find the current passing through the entire element to calculate current ,we need to sum up all the resistances in the circuit,, eliminating the parallel circuit in the network R34= R3,R4 are in series = 390 ohms R2, R5,R34 are in series its equivalent is 1/ R2345 = 1/R2+1/R5+1/R34 =1/114+ 1/213 +1/ 390 R2345 =62.3 ohms next R6,R7 are in parallel 1/R67 = 1/R6+1/R7 =1/293+1/260 R67= 137.75 ohms so R1,R2345 ,R67 ,R8 are in series and gives the equivalent resistance of the entire network Requ = 94+ 62.3+ 137.75 +207= 501.05 ohms =501 ohms I =V/Requ =26/501 =0.051 amp =51 milli amp= 51ma A) V= IR V1 =I R1 V1= 4.79 V note: in parellel ckt ,,voltage will be equal across all its branches and current will divided so voltage across R2,R34,R5 are equal V2345=IR2345 = 51ma *62.3 = 3.17 V so V2= V5 = 3.17 V V3= 1.99V V4= 1.18V V6 =V7 =7.025V V8 = IR8 = 10.557 so final answ for A bit is 4.79,3.17,1.99,1.18,3.17,1.025,10.557 B) I1 =51ma I2=27ma I3=8.1ma I4=8.1ma I5= 27ma I6=12.1ma I7=3.4ma I8=51ma

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