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(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
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 =
I3 = I1*Y3 = 0.1024*0.6285 = 0.064
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
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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