Solving for current in circuit with several light bulbs

In summary, the given problem involves a circuit with a voltage source of 20V and three bulbs with voltages of 7.0V, 5.0V, and 8.0V. The total current is 150 mA, with one bulb having a current of 40 mA. The task is to find the currents for the remaining two bulbs. The solution involves applying Kirchoff's loop rule and KCL, and using conventional current in calculations.
  • #1
Derpicus
8
3

Homework Statement


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Homework Equations


Series[/B]
Vt= V1 + V2 + V3 etc
It = I1 = I2 = I3 etc
Parallel
Vt = V1 = V2 = V3 etc
It = I1 + I2 + I3

The Attempt at a Solution



Vt= 20V V2= 5.0 V V4= 8.0V
It= 150 mA I2= 40 mA
Find I3, and I4

I've solved for the values of V already
V1 = 7.0V
V3 = 5.0V
So Vt = V1+V2=V3+V4
= 7.0V + 5.0V + 8.0V = 20V
But I'm getting confused on solving for I which is current.

I1 is in series with I2 and I4, but I2 is also parallel to I3. So which set of laws would I apply for solving for I3?
I know I1=I4 since they're in series.
I'm assuming I1=I1=It - I2
= 150 mA - 40mA = 110mA
But still stuck on how to solve for I3, since its not stated the bulbs are identical so I'm unsure if I3 would also be 40mA. I'd appreciate any help I can receive.
 
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  • #2
Shouldn't current be in the opposite direction?
Also, Have you tried applying Kirchoff's loop rule?
 
  • #3
Titan97 said:
Shouldn't current be in the opposite direction?
Also, Have you tried applying Kirchoff's loop rule?
I copied everything from the diagram the current flows the way it does since it starts from the negative part of the cell and flows along to the positive part of the cell. I also don't believe I've learned about Kirchoff's loop rule yet. Once it reaches the box where I2 is it has two possible paths going down to V2 or proceeding forward and then down to I3/V3.
 
  • #4
Derpicus said:
I copied everything from the diagram the current flows the way it does since it starts from the negative part of the cell and flows along to the positive part of the cell. I also don't believe I've learned about Kirchoff's loop rule yet. Once it reaches the box where I2 is it has two possible paths going down to V2 or proceeding forward and then down to I3/V3.
The direction of current is conventionally from +ve terminal to -ve terminal of the battery, as Titan97 said earlier. But using the given convention, it won't affect the calculations once you flip the voltage polarities accordingly. Also, there is only one source present(Vt), rest all are bulbs.
Where is It in your diagram?
 
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  • #5
cnh1995 said:
The direction of current is conventionally from +ve terminal to -ve terminal of the battery, as Titan97 said earlier. But using the given convention, it won't affect the calculations once you flip the voltage polarities accordingly. Also, there is only one source present(Vt), rest all are bulbs.
Where is It in your diagram?
Itotal is just past V4
 
  • #6
Derpicus said:
Itotal is just past V4
Then isn't It=I1? You have I1 and I2. What is the relation between i1, i2 and i3?
 
  • #7
cnh1995 said:
Then isn't It=I1? You have I1 and I2. What is the relation between i1, i2 and i3?
I know I1 is 110mA
Since I2 is 40mA
As I1=I4=It-I2
Since the total incoming has to equal the total outgoing.

But the issue I'm having is I3 is parallel to I2

Since 110ma+40 mA = 150 mA which is the It.

But the relationship for parallel is It= I1+I2 etc, so I'm unsure how to go about solving for I3.
 
  • #8
Derpicus said:
As I1=I4=It-I2
If It is just past V4, then I1=I4=It.
 
  • #9
cnh1995 said:
If It is just past V4, then I1=I4=It.
KCL says incoming current at a point is equal to the outgoing current. If you apply that, you can see
I1=I2+I3.
 
  • #10
Where are the several sources? It looks like only one. Shouldn't all V's be the same if there are no resistors?
 
  • #11
justaman0000 said:
Where are the several sources? It looks like only one. Shouldn't all V's be the same if there are no resistors?
There appears to be only one source.

No. The V-s are not all the same. Several are given.

Apparently the problem statement is that:
Vt= 20V, V2 = 5.0 V, V4= 8.0V
It= 150 mA, I2= 40 mA

Find I3, and I4​

And it looks like OP uses electron current, rather than conventional current..
 
  • #12
SammyS said:
There appears to be only one source.

No. The V-s are not all the same. Several are given.

Apparently the problem statement is that:
Vt= 20V, V2 = 5.0 V, V4= 8.0V
It= 150 mA, I2= 40 mA

Find I3, and I4​

And it looks like OP uses electron current, rather than conventional current..
Thank you SammyS
 

1. How do I calculate the current in a circuit with multiple light bulbs?

The current in a circuit with multiple light bulbs can be calculated by using Ohm's law, which states that current (I) is equal to the voltage (V) divided by the resistance (R). In a series circuit, the total resistance can be found by adding up the individual resistances of each component. Once the total resistance is known, the current can be calculated by dividing the voltage by the total resistance.

2. What is the difference between series and parallel circuits?

In a series circuit, all components are connected in a single loop, meaning the current flows through each component in succession. In a parallel circuit, the components are connected in multiple branches, allowing the current to flow through each component simultaneously. This results in different calculations for current and total resistance in each type of circuit.

3. How do I calculate the total resistance in a parallel circuit?

In a parallel circuit, the total resistance can be calculated by using the formula 1/Rt = 1/R1 + 1/R2 + 1/R3 + ..., where Rt is the total resistance and R1, R2, R3, etc. are the individual resistances of each component. Once the total resistance is known, the current can be calculated using Ohm's law.

4. How does the brightness of a light bulb affect the current in a circuit?

The brightness of a light bulb is directly related to the current flowing through it. The higher the current, the brighter the bulb will be. This is because the brightness of a bulb is dependent on the amount of energy (in the form of electricity) that is being converted to light. Therefore, increasing or decreasing the current will also affect the brightness of the bulb.

5. Can I use the same calculations for current in a DC and AC circuit?

Yes, the same calculations can be used for current in both DC (direct current) and AC (alternating current) circuits. However, in AC circuits, the current is constantly changing direction, so the calculations may need to be adjusted to account for this. Additionally, the units of measurement for AC current are typically given in amperes (A) or milliamperes (mA), while DC current is typically measured in amperes or microamperes (μA).

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