Three 11.0 resistors are connected in series with a 15.0 V battery. Find the fol

Question

Three 11.0 resistors are connected in series with a 15.0 V battery. Find the following. (a) the equivalentresistance of the circuit
Enter anumber. 1

(b) the current in each resistor
Enter anumber. 2 A

(c) Repeat for the case in which all three resistors are connectedin parallel across the battery. equivalent resistance Enter anumber. 3 current in each resistor Enter anumber. 4 A Enter anumber. Enter anumber. Enter anumber. Enter anumber. equivalent resistance Enter anumber. 3 current in each resistor Enter anumber. 4 A

Explanation / Answer

3 resistor in series, equivalent resistance is the sum ofthe resistors: R(1)+R(2)+R(3) = 11+11+11 = 33 Current is found with Ohms Law: V = I*R Rearrange for I: I = V/R I = 15/33 I = 0.45 amps (series circuit current is constant, so currentin each resistor is the same) For the parallel equivalent resistance, since all resistorsare the same value: R/N      ----> This equation isvalid for any number of parallel resistors of EQUAL VALUE where R is the resistor value and N is the number of parallelresistors R(eq) = 11/3 = 3.67 To find the total circuit current, use Ohm's Law again: I = V/R I = 15/3.67 I = 4.09 amps Since the sum of individual branch currents in a parallelcircuit equals the total current, and all branch resistances arethe same, then the current splits equally: 4.09 amps / 3 branches (each of 11 ohms) = 1.36 amps per branch Hope this helps.

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