# Solve Open Circuit Problem 3: Two Batteries and 3 Bulbs

• gtqueen371
In summary, an open circuit problem is when there is a break or interruption in the connection of a circuit, preventing the flow of electricity. In the specific scenario of using two batteries and 3 bulbs connected in series, any break in the connection will result in an open circuit. To solve this problem, one must check and fix any faulty connections and ensure all components are properly connected and have enough power. Open circuit problems can be caused by various factors, such as loose connections, damaged components, and low battery power. To prevent these problems, regular maintenance, use of high-quality components, and protection from environmental factors are recommended.
gtqueen371
Three bulbs
A circuit is made of two 1.5 volt batteries and three light bulbs as shown in Figure 19.57. When the switch is closed and the bulbs are glowing, bulb 1 has a resistance of 9 ohms, bulb 2 has a resistance of 44 ohms, bulb 3 has a resistance of 28 ohms, and the copper connecting wires have negligible resistance. You can also neglect the internal resistance of the batteries.

Picture Attached of Circuit

2. (a) With the switch open, indicate the approximate surface charge on the circuit diagram. Which of the following statements about the circuit (with the switch open) are true:

1. The electric field in the filament of bulb 3 is zero.
2. There is a large gradient of surface charge between locations M and L.
3. The surface charge on the wire at location B is positive.
4. The electric field in the air between locations B and C is zero.
5. There is no excess charge on the surface of the wire at location C.

I chose 2, 3, and 4. This is not the correct combination of answers. 1 should be wrong because there is current flowing through the filament, E can't be xero in the bulb. I figured 2 was correct because the wire would have a large negative gradient as current leaves the negative end of the battery. I thought 3 would be correct due to it being on the side of the circuit closest to the + end of the battery, and I picked 4 because I thought the charges on either side of the switch would cancel each other out. There is a negative excess surface charge at C, so 5 is incorrect.

(b) With the switch open, find these potential differences:
1. VB - VC =
2. VD - VK =

I thought 1 was 0 because current isn't traveling past the switch since its open, and I assumed 2 was 3 because this the same as the delta V leaving the battery.

(c) After the switch is closed and the steady state is established, the currents through bulbs 1, 2, and 3 are I1, I2, and I3 respectively. Which of the following equations are correct loop or node equations for this steady state circuit?
1. I1 = I2 + I3
2. -I1*(9 )-I2*(44 ) + I3*(28 ) = 0
3. -I2*(44 ) + I3*(28 ) = 0
4. + 3V + -I1*(9 ) + -I2*(44 ) = 0
5. I2 = I3
6. +3V + -I1*(9 ) + I3*(9 ) = 0

I chose 1, 4, and 5. Perhaps I should have also chosen 2?

(d) In the steady state (switch closed), which of these are correct?
1. VC - VF = +I1*(9 )
2. VC - VF = +I2*(44 )
3. VC - VF = 0
4. VL - VA = -3V + I1*(9 )
5. VC - VF = +I3*(28 )

I chose 2 and 4.

(f) Now find the unknown currents, to the nearest milliampere.
I1 =
I2 =
I3 =

I doesn't matter what I got here because the calculations were wrong due to earlier mistakes.

(g) How many electrons leave the battery at location N every second?
____________electrons/s

Incorrect calculation due to earlier assumptions, but i=I/q.

(i) What is the numerical value of the power delivered by the batteries?
P = ___________W

Incorrect calculation due to earlier assumptions, but P=deltaV/R.

(j) The tungsten filament in the 44 ohm bulb is 10 mm long and has a cross-sectional area of 2 multiplied by 10-10 m2. What is the magnitude of the electric field inside this metal filament?
|e vector| = ________ V/m

Incorrect calculation due to earlier assumptions, but E=emf/L.

I am a first time poster, tyring out this site, and any help will be GREATLY appreciated! :0)

#### Attachments

• circuit.gif
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Try and ignore all the traps making you do parts a) b) and c) first.
Lets go for 2d, because it will give you a testbed on which to try out the options. We can do them later.

Take a few general principles. While the switch is open, there is no current. No electrons are flowing. It is true that when the circuit was constructed, some electrons flowed briefly to put a charge in all the bits, but then it stopped. If you ever manage to put a voltage across a bulb resistance, there will be a current through it. So - with no current, what would you expect if you tried to measure a voltage across a bulb?

If you have learned to figure series and parallel resistances, work out the total resistance in the circuit.
If you cannot, then post here what you think is the way to calculate the total of 2 resistances in parallel.

If you are OK so far, then use the voltage in the circuit, and the total resistance you just found, to find the main current in the circuit if you close the switch (I3).

I3 is divided into some that goes through bulbs (2) and (3) unequally, but you can get to it by knowing that the thing they have in common is the voltage across them. So press on. Now you know the current, use it to also find the voltages dropped across the resistances.

By now, you are getting to know quite a lot. You now have the means to apply the very same law to figure I1 and I2. if you get stuck, then post here your attempt.

Once we have this part untangled, we can better see how do do the other bits. You may find it starts to occur to you.

I don't know how to get the last question

(j) The tungsten filament in the 44 ohm bulb is 10 mm long and has a cross-sectional area of 2 multiplied by 10-10 m2. What is the magnitude of the electric field inside this metal filament?

can anyone answer the last question for me??

I'm sorry, I have tried going back and understanding what has been posted but I still can't figure out part (a) of this problem

Welcome to Physics Forums.
I'm sorry, I have tried going back and understanding what has been posted but I still can't figure out part (a) of this problem
It is forum policy that students show an attempt at solving a problem before getting help. With that in mind, what are your thoughts on the various answers given in part (a), and why?

## What is an open circuit problem?

An open circuit problem occurs when there is a break or interruption in the connection of a circuit, preventing the flow of electricity.

## How do two batteries and 3 bulbs create an open circuit problem?

In this specific scenario, the two batteries and 3 bulbs are connected in series, meaning they are connected end to end. If there is a break in the connection at any point, the circuit will be open and the bulbs will not light up.

## How can I solve an open circuit problem with two batteries and 3 bulbs?

To solve this problem, you will need to check the connections between the batteries and the bulbs. If there is a break, you will need to fix it or replace the faulty component. It is also important to ensure that all components are properly connected and that the batteries have enough power to light up the bulbs.

## What are some common causes of open circuit problems?

Open circuit problems can be caused by a variety of factors, including loose or faulty connections, damaged components, and low battery power. Environmental factors such as extreme temperatures or moisture can also contribute to open circuit problems.

## How can I prevent open circuit problems in the future?

To prevent open circuit problems, it is important to regularly check and maintain all connections and components in a circuit. It is also important to use high-quality and properly rated components, as well as protect the circuit from environmental factors. Properly shutting off and storing batteries when not in use can also help prevent open circuit problems.

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