# Solving for I1/I2 with Current Divider Rule: Help Needed!

• freshbox
In summary, the conversation is about finding the value of I1/I2 using the Current Divider Rule. The person is unable to solve it and is seeking help. They share their working and it is pointed out that there was an error in adding the resistor. The person also asks about the voltage of the capacitor and it is explained that at steady state, the capacitor acts like an open circuit and has a maximum voltage. The conversation ends with the person understanding their mistake and asking about the voltage drop across the resistor when the current is zero.
freshbox
Reference to the below 1st picture, I found that I = 0.916A.
I am trying to find I1/I2 and I tried to used the Current Divider Rule (2nd picture) But I am unable to solve it. May I know why?

Thanks.

#### Attachments

• rlc.jpg
20.7 KB · Views: 402
• 2nd.jpg
26.2 KB · Views: 368
freshbox said:
Reference to the below 1st picture, I found that I = 0.916A.
I am trying to find I1/I2 and I tried to used the Current Divider Rule (2nd picture) But I am unable to solve it. May I know why?

Thanks.

You'll have to show your attempt so that we can see how to help.

The first image is very dark; What's the value of the battery voltage? Is it 11 V?

I'm sorry. Yes the battery voltage is 11v.

My working:
Since I=0.916A

Using current divider rule to find i1:
I1=I x R3/R2+R3
I1=0.916 x 2
=0.916 x 2
=1.832A (Wrong Ans)

However the actual answer is 0.46A

So I'm wondering how come I cannot use CDR to solve for i1 or i2.

freshbox said:
I'm sorry. Yes the battery voltage is 11v.

My working:
Since I=0.916A

Using current divider rule to find i1:
I1=I x R3/(R2+R3) <--- use parentheses to make operations clear!
I1=0.916 x 2 <---- How did you arrive at 2? What are R2 and R3?

Your approach is okay, but I think you've slipped up on plugging the correct values into the current divider equation.

I suspect that there is an inductor (L2) below resistor (R2) that's why I cannot use CDR.

freshbox said:
I suspect that there is an inductor (L2) below resistor (R2) that's why I cannot use CDR.

Nope. No problem there; the inductor "disappears" at steady state. The problem seem to be what you're plugging in for R2 and R3, or if not that, what value you arrive at when you do the math.

I have another question to ask you: Is the voltage for the Capacitor is 10v or some of the voltage are absorb by by the Resistor?

To my understanding, at 5T (Steady State), Capacitor acts like an open circuit hence there is no current but it's voltage is at max value.

But I'm not sure whether the Capacitor is having full voltage or some of the voltage is absorb by the R (2k ohm)

Thank you.

#### Attachments

• t.jpg
33.8 KB · Views: 347
freshbox said:

I have another question to ask you: Is the voltage for the Capacitor is 10v or some of the voltage are absorb by by the Resistor?

To my understanding, at 5T (Steady State), Capacitor acts like an open circuit hence there is no current but it's voltage is at max value.

But I'm not sure whether the Capacitor is having full voltage or some of the voltage is absorb by the R (2k ohm)

Thank you.

After a long time (5T and more), the current will go to zero. What's the voltage drop across the resistor if the current is zero?

0v. Capacitor = 10v

## 1. How do I use the current divider rule to solve for I1/I2?

The current divider rule states that the current flowing through any branch of a parallel circuit is inversely proportional to the resistance of that branch. To solve for I1/I2, you would first calculate the total resistance of the circuit, then use the formula I1/I2 = R2/(R1+R2).

## 2. What is the purpose of using the current divider rule?

The current divider rule is used to determine the individual currents flowing through each branch of a parallel circuit. It helps in analyzing and understanding the behavior of the circuit.

## 3. Can I use the current divider rule for any type of circuit?

The current divider rule is applicable to parallel circuits only. It cannot be used for series or combination circuits.

## 4. How do I know if I have applied the current divider rule correctly?

To ensure that you have applied the current divider rule correctly, you can check if the sum of the individual currents through each branch equals the total current of the circuit. If it does, then you have applied the rule correctly.

## 5. Are there any limitations to using the current divider rule?

Yes, the current divider rule assumes that the branches of the parallel circuit are ideal and have equal voltage sources. In real-world circuits, this may not always be the case, so the rule may not provide an accurate result.

Replies
9
Views
2K
Replies
9
Views
2K
Replies
10
Views
1K
Replies
8
Views
1K
Replies
1
Views
939
Replies
4
Views
3K
Replies
18
Views
2K
Replies
6
Views
2K
Replies
4
Views
1K
Replies
3
Views
1K