# A circuit is made of two 1.4 volt batteries and three light bulbs as shown in... A circuit is made of two 1.4 volt batteries and three light bulbs as shown in the figure. When the switch is closed and the buibs are glowing, bulb I hasa resistance of 11 ohms, bulb 2 has a resistance of 44 ohms, bulb 3 has a resistance of 26 ohms, and the copper connecting wires have negligible resistance. You can also neglect the internal resistance of the batteries (a) With the swtch open, indicate the approximate surface charge on the cirouit ciagram.(Do this on paper. Your instructor may ask you to turn in this work.) Refer to your diagram to decide which of the following statements about the circuit (with the switch open) are There is a large gradient of surface charge between locations M and L The surface charge on the wire at location B is positive. The electric field in the air between locations B and C is zero. The electric field in the filament of bulb 3 is zero. There is no excess charge on the surface of the wire at location C. (b) With the switch open, find these potential differences: (c) After the switch is closed and the steady state is established, the currents through bubs 1, 2, and 3 are I. Iz, and Jy respectively. Which of the followling equations are correct loop or node equations for this steady state circuit? (d) in the steedy state (switch closed). which of these are correct? T) Now find the unknown currents, to the nearest milliampere. (T.e. enter your answer to three decimmal places (9) How many electrons leave the battery at location N every second? What is the numerical value of the power delivered by the U) The tungsten filament in the 44 ohm bulb is 5 mm long and has a cross-sectional area of 2.101 m2. What is the magnitude of the electric field inside this metal filament?

(b) When the switch is open the current through the circuit is zero. So, total voltage dropped across the open circuited switch.

VB - V C = V = 1.4+1.4

VB - V C = 2.8 Volts

(f) You can use resistances to solve this problem

I1 is the master current. It will equal the master voltage from the batteries, divided by the total circuit equivalent resistance.

To get the total circuit equivalent resistance, we first combine R2 and R3 in parallel to get R23. Then we combine R23 in series with R1.

Rnet = R1 + R2*R3/(R2 + R3)

Note: use the "product over sum" shortcut with caution. This is only applicable for two branches of resistances.

I1 = Vbat / Rnet
I1 = 2 * Vbat/Rnet
I1 = 2*Vbat / (R1 + R2*R3/(R2 + R3))

I1 = 2*1.4/ (11 + [(44*26) / (44 + 26) ] )

I1 = 0.1024 A

Now, divide this current up, according to the CONDUCTANCE fractions of both paths. The conductances are the reciprocals of resistances. The conductance fractions are the fractions of the total conductance.

Total conductance:
1/R2 + 1/R3

Conductance fraction of path through R2:
Y2 = (1/R2)/(1/R2 + 1/R3)
Y2 = 0.3714

Conductance fraction of path through R3:
Y3 = (1/R3)/(1/R2 + 1/R3)
Y3 = 0.6285

I2 = I1*Y2 = 0.1024*0.3714 = 0.038
I3 = I1*Y3 = 0.1024*0.6285 = 0.064

Results:
I1 = 0.102 A
I2 = 0.038 A
I3 = 0.064 A

(g) The current flowing at location N, is I1. It is flowing in to the battery, thus electrons are exiting the battery. We need to translate from amperes per second, into elementary charge units per second, since the charge of 1 electron is 1 elementary charge unit. 1 elementary charge unit is 1.602 * 10-19 Coulombs. And the Ampere is 1 Coulomb per second. Believe it or not, the ampere came first in the original definitions.

Divide by the elementary charge unit, in Coulombs
0.102 A / 1.602 * 10-19 C  = 6.367*1017 electrons/second

(i) Multiply the master current and voltage to get the total power.
P = I1*Vbat = 0.102 * 2.8

P = 0.2856 Watts

( j ) Electric field is voltage divided length, assuming that it is uniform. In a wire such as this, we might as well assume it is uniform. The cross sectional area doesn't really matter, all that does is assist us in confirming that it is 44 ohms.

E = V/L
V = I2*R2
E = I2*R2/L
E = (0.038 A)*(44 ohms)/(0.005 m)

|E| = 334.4 V/m

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