Question

You have a circuit that has 4 resistors and a 24 volt battery Let the resistors have resistances of R 1 =4Ω,R 2​=6Ω,R 3 =8Ω,R 4 =10Ω. Answer the following questions using Kirchoff's rules:

(a) What is the current through R 3 in a circuit where you have the battery, then the first two resistors (R 1,R 2) hooked up in series, followed by the second two resistors (R 3,R 4 ) hooked up in parallel, before reconnecting back to the battery. (Show your circuit and what you choose for the loop(s) and junction(s) as part of your solution).

Answer 1

The current through resistor R3, when R1 and R2 are in series and R3 and R4 are in parallel, is calculated using Ohm's law. After computing the equivalent resistance of the parallel and series combinations, the current through R3 is found to be 3 A.

Step-by-step explanation:

To find the current through R3 in a circuit where R1 = 4 Ω and R2 = 6 Ω are connected in series, and R3 = 8 Ω and R4 = 10 Ω are connected in parallel, we can use Kirchhoff's rules.

First, we find the equivalent resistance of R3 and R4 in parallel:

R34 = 1 / ((1/R3) + (1/R4)) = 1 / ((1/8) + (1/10)) = 4.44 Ω

The total resistance in the circuit is the sum of the series resistances: Req = R1 + R2 + R34 = 4 Ω + 6 Ω + 4.44 Ω = 14.44 Ω.

Using Ohm's law, V = I * R, the total current supplied by the battery is: I = V / Req = 24 V / 14.44 Ω = 1.66 A.

The voltage drop across R1 and R2 is the same as the total voltage supplied across R34, which is 24 V. Hence, using Ohm's law again, the current through R3 can be determined: I3 = V / R3 = 24 V / 8 Ω = 3 A.