Compute the equivalent resistance of the network, and find thecurrent in each re

Question

Compute the equivalent resistance of the network, and find thecurrent in each resistor. (R1 = 3.00 , R2 = 9.00 , R3 = 3.00 , R4 = 5.00 , and = 53.0 V.) The battery has negligible internalresistance.
equivalent resistance ??? current in each resistor ??? A (Resistor 1) ??? A (Resistor 2) ??? A (Resistor 3) ??? A (Resistor 4) Compute the equivalent resistance of the network, and find thecurrent in each resistor. (R1 = 3.00 , R2 = 9.00 , R3 = 3.00 , R4 = 5.00 , and = 53.0 V.) The battery has negligible internalresistance.
equivalent resistance ??? current in each resistor ??? A (Resistor 1) ??? A (Resistor 2) ??? A (Resistor 3) ??? A (Resistor 4)

Explanation / Answer

R2 and R4 are in series .So, resultant R ‘ = R 2+ R4 = 9+ 5 = 14 ohm

R and R ‘ are in parallel.So, equivalent resistance R“ = ( R*R’) / (R+R’)

               R “ = 4.2 ohm

Current through the circuit I = E / R “

                                               = 53 V / 4.2 ohm

                                               = 12.619 A

Current through R is I’ = I * R ‘ / ( R + R ‘)

                                     = 8.833 A

Current through R ‘ is I “ = I * R / ( R + R ‘)

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