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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. V_{B}- V_{C }=

2. V_{D}- V_{K}=

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)

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# Homework Help: Open Circuit Problem

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