Finding the current and the voltage across each resistor

[Lopez]

Homework Statement

Find the current and the voltage across each resistor.

volts is 24v
R1=270 ohms
R2=330 ohms
R2=1.2k ohms
R4=1.2k ohms

Homework Equations

The Attempt at a Solution

This is not a homework its more like a study guide.
He gave us the answer for just the currents which are
IR1=20
IR2=20
IR3=10
IR4=10
But not sure how he got that.
this is what i did but don't know how to go from there:
(270+330)|| (1.2k||1.2k)
600||600
R=300

 

[Chestermiller]

Hi Lopez. Welcome to physics forums!

The 270 and 330 are in series with the 600, not parallel.

Chet

 

[haruspex]

(270+330)|| (1.2k||1.2k)

Having reduced the (1.2k||1.2k) to an effective 0.6k, what is the relationship of that in the circuit to R1 and R2?

 

[Lopez]

ahh yes there was the problem,after 1.2 k and the other 1.2k were combined it becomes one resistor in series with the (r1 and R2). Thanks!

 

[Nickie]

ohms
V=24V
I=V/R
I=24/300
I=0.08A or 80mA

Based on the given information, it appears that the circuit is a series-parallel circuit with two parallel branches. To find the current and voltage across each resistor, we can use Ohm's Law (V=IR) and Kirchhoff's Laws.

First, we can calculate the total resistance of the circuit by using the parallel resistance equation: 1/Rt = 1/R1 + 1/R2 + 1/R3 + 1/R4

1/Rt = 1/270 + 1/330 + 1/1200 + 1/1200
1/Rt = 0.0069 + 0.003 + 0.00083 + 0.00083
1/Rt = 0.01146
Rt = 1/0.01146
Rt = 87.2 ohms

Next, we can calculate the total current in the circuit by using Ohm's Law: I = V/R

I = 24/87.2
I = 0.275A or 275mA

Now, we can use Kirchhoff's Current Law (KCL) to find the current through each resistor. KCL states that the sum of the currents entering a node (or junction) must equal the sum of the currents leaving the node. In this case, the current entering the node is the total current (0.275A) and the currents leaving the node are the individual currents through each resistor.

Using KCL for the left node:

0.275A = IR1 + IR2
IR1 + IR2 = 0.275A

Using Ohm's Law, we can solve for IR1 and IR2:

IR1 = V/R1
IR1 = 24/270
IR1 = 0.0889A or 88.9mA

IR2 = V/R2
IR2 = 24/330
IR2 = 0.0727A or 72.7mA

Using KCL for the right node:

0.275A = IR3 + IR4
IR3 + IR4 = 0.275A

Using Ohm's Law, we can solve for IR3 and IR4:

IR3 = V/R3
IR3