Question

11)

For the circuit shown, R₁= R₃ = R₉ = R₁₁= 102.0 Ohms, the rest of the resistors are 204.0 Ohms.

a) Without doing any simplification (or combination) identify the number of pairs of resistors (that is 2 resistors) that are in series.

b) Without doing any simplification (or combination) identify the number of pairs of resistors (that is 2 resistors) that are in parallel.

c) Find R₁₂ the equivalent resistance of resistors R₁, and R₂.

d) Find R₁₂₃ the equivalent resistance of resistors R₁, R₂, and R₃.

e) Find R₁₂₃₄₅ the equivalent resistance of resistors R₁, R₂, R₃, R₄, and R₅.

12)

Consider the circuit shown. All light bulbs are identical with a resistance of 6.0 Ohms. The battery is V=6.0V:

a) What is in ? the resistance seen by the battery?

b) What is in A the current through bulb C?

c) What is in A the current through bulb A?

d) What is in V the voltage across bulb D?

e) What is in W the power dissipated in bulb H?

13)

Consider the circuit shown for which R₁=1.3 M?, R₂=0.8 M?, R₃=2.1 M?, R₄=2.9 M?, C=4.8 ?F,and E=19.0V. For charging, we open S₂ and close S₁. For discharge, after a full charge, we open S₁ then close S₂.

a) What is the charge time constant for the circuit? Answer in units of seconds.

b) What is the discharge time constant for the circuit? Answer in units of seconds.

After fully discharging the circuit, we close S₂ and then close S₁. The capacitor will charge.

c) What is in micro-amps the current through R₂ once the capacitor is fully charged?

d) What is in micro-amps the current through R₃ once the capacitor is fully charged?

14)

For the wires shown in the Figure below, the wires on the left carry a current of I₁ = I₂ = 1.3A (out of the page), the wire on the right carries a current of I₃ = 3.9A (into the page), a = 0.17 m and b= 0.36 m.

a) Find in Tesla the magnitude of the magnetic fields at point X due to the wire carrying I₃.

b) Find the direction (as an angle relative to traditional positive x-direction) of the magnetic fields at point X due to the wire carrying I₃.

c) Find in Tesla the magnitude of the magnetic fields at point X due to the wire carrying I₁.

d) Find the direction (as an angle relative to traditional positive x-direction) of the magnetic fields at point X due to the wire carrying I₁.

e) Find in Teslas the magnitude of the overall magnetic field due to all 3 wires at point X.

15)

Consider a mass spectrometer like the one shown in the figure. A beam of singly charged ions (charge is 1.6 x 10^-19 C) is accelerated from an ion source and injected into a velocity selector with an electric field of E=2400V/m) and magnetic field of B=43.0mT. The ions then enter the deflection region where a uniform B₀=31.0mT magnetic field is maintained. As shown in the Figure, the ions are observed to strike a detector at a distance of 0.505m from the entrance of the deflection chamber.

a) What would be in m/s the velocity of the ions that move in a straight horizontal line when moving through the velocity selector?

b) What would be in meters the radius of the path of the ions when moving through the Deflection Chamber if they hit the detector at a distance of 0.505m from the entrance of the deflection chamber?

c) What is the sign of the charge of the ions if they follow the path shown in the Figure?
negative   positive    both negative and positive ions would follow that same path

d) What is in kg the mass of the ions that hit the detector  at a distance of 0.505m from the entrance of the deflection chamber?

Answer 1

15 a) 55.814 *10^3 m/s

b) 0.2525 m

c) positive

d) 2.244 * 10^-26 kg

Answer 2

a)only, R1 and R2

b) R6 and R7

R10 and R11

c) R12 = R1 + R2

= 204 + 204

= 408 ohms

d) R123 = R12*R3/(R1+R3)

= 408*102/(408+102)

= 81.6 ohms

e) R12345 = R123*(R4+R5)/(R123+R4+R5)

= 81.6*(204+204)/(81.6+204+204)

= 68 ohms

Answer 3

istance on sfnphfafion 12 qv 几 58 12 2 /2 S8 a) yestance Seen by baffexy 1055)

= o.isshq. a) refhnce Seen by b.ffexy 3 b) Curent though C%; oISSRmp /2 +12 Yp= 1D-12=0.620689 Volfs.

Answer 4

a) Apply, q*v*B = q*E

v = E/B

= 2400/(43*10^-3)

= 5.58*10^4 m/s

b) r = d/2

= 0.505/2

= 0.0.2525 m

c) positive

d) Apply, m*v^2/r = q*v*B

m = B*q*r/v

= 31*10^-3*1.6*10^-19*0.2525/5.58*10^4

= 2.24*10^-26 kg

Answer 5

a) B3 = mue*I3/(2*pi*a)

= 4*pi*10^-7*3.9/(2*pi*0.17)

= 4.59*10^-6 T <<<<<<<----------Answer

b) direction of B3 is towards +x axis <<<<<<<----------Answer

c) B1 = mue*I1/(2*pi*sqrt(a^2+b^2))

= 4*pi*10^-7*1.3/(2*pi*sqrt(0.17^2+0.36^2))

= 6.53*10^-7 T <<<<<<<----------Answer

d) theta = tan^-1(a/b) + 90

= tan^-1(0.17/0.36) + 90

= 25.3 + 90

= 115.3 degrees with +x axis <<<<<<<----------Answer

e) B2 = mue*I2/(2*pi*b)

= 4*pi*10^-7*1.3/(2*pi*0.36)

= 7.2*10^-7 T (towards +y axis)

B1x = 6.53*10^-7*cos(115.3) = -2.79*10^-7 T

B1y = 6.53*10^-7*sin(115.3) = 5.9*10^-7 T

B2x = 0

B2y = 7.2*10^-7 T

B3x = 4.59*10^-6 T

B3y = 0

Bnetx = B1x + B2x + B3x = 4.311*10^-6 T

Bnety = B1y + B2y + B3y = 1.31*10^-6 T

Bnet = sqrt(Bnetx^2 + Bnety^2)

= sqrt(4.311^2+1.31^2)

= 4.5*10^-6 T <<<<<<<----------Answer