Consider the following circuit: Let R1 = 5, R2 = 10 Ohms and R3 = 5 Ohm. Let Vl

Question

Consider the following circuit: Let R1 = 5, R2 = 10 Ohms and R3 = 5 Ohm. Let Vl = 20 V and V2 = 12V. Using Kirchoffs rules, find the currents II, 12, and 13 flowing in the circuit Find the power dissipated in each resistor and the total power dissipated in the circuit. Consider the following circuit We initially close the switch to charge the capacitor. Find the charge Qo built up on the capacitor, as well as the current lo in the resistor. At time t=0, we open the switch thus disconnecting the voltage source from the electric circuit. Write down the charge Q(t) as a function of time t as well as the current l(t) as a function of time. Draw Q(t) and l(t) as a function of timeA rectangular loop of wire of length = 10 cm, supporting a mass m = 1 gram, hangs vertically with one end in a uniform magnetic field B = 100 Gauss. For what current in the loop would the magnetic field force upward exactly balance the gravitational force downward? What is the direction of the current (clockwise or counter clockwise?) only the top segment of the loop will experience an upwards force. A point particle q that has charge = 4.5 xlO'6 C and mass 10 5 Kg. moves at a velocity = 3 m/s x along the line y=3m in the x-y plane as shown. A magnetic field = 1 Tesla is turned on in the y direction , B = 1 /. T Assume at time t = 0, the charge q is at x = 0, y = 3 m and z = 0. What is the direction of the force on the particle? What is the magnitude of the force on the particle? What is the acceleration of the particle in the direction of the force?

Explanation / Answer

Q1 Ans:-Part a by using Kirchhoff’s law

At point c

I3 = I1 +I2…………..(1)

In 1st loop

V1 – R1I1 – R3I3=0

V1 = R1I1 +R3I3

20=5I1+5I3…………….(2)

In 2nd loop

V2 = I2R2+ I3R3

12= 10I2+ 5I3………………..(3)

By using total loop

V1 – I1R1+ I2R2 –V2 =0

8 =5I1- 10I2………………………………..(4)

Put equ 1 in equ 2

20= 10I1+5I2………………….(5)

Now eq 5*2 +eq 4 we get

I1= 1.9A

Eq 1 put in eq 3

12 = 15I2 + 5I1 ……………….(6)

Eq 6 – eq 4 we get

I2 = 0.16A

I3 = 2.08 A

Part b :-

P1=I1^2*R1 = (1.9)^2 *5 =18.05w

P2= 0.16^2*10= 0.256w

P total power =(2.08)^2*5= 21.63w

Q2 Ans:-Part a

Q0= V*C= 12*1*10^-6 =12*10^-6c

I0=dq/dt=(/R)*e^(-t/R*C)=12/100*1 =012A

Part b:-

Q(t) = C*(1- e^(-t/R*C))

I(t)= dQ/dt= /R*(e^(-t/R*C))

Q3 Ans:-

Here magnetic force and gravitational force balance each other so

Fb =Fg

I*L * B* sin = M*g

I *10*10^-2*100 *sin90 = 1*10^-3*9.8

I=9.8*10^-4A……………………counter clockwise direction

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