Electric Current Exam1 and  Problem Solutions

1. Voltage vs. current graph of a conductor is given below. Find the change in the resistance of conductor in first and third intervals.

ohmslaw_1

 

 

 

 

 

 

 

 

We use ohm's law to find relation between V, I and R.

Interval I: Since potential and current increase linearly, resistance of the conductor becomes constant.

Interval II: In this interval, potential is constant but current increases. Thus, resistance of the conductor must decrease to make potential constant.

Interval III: In this interval potential increases but current is constant. Thus, resistance of the conductor must increase to make potential increase.

 

2. a)Find the equivalent resistance of the circuit given below.

b)Find current passing through the circuit.

 ohmslaw_2

 

 

 

 

 

 

 

 

a) Since resistance are connected in parallel between points A and B;

ohmslaw_2_solution

 

 

 

 

 

 

1/Req=1/4+1/4

Req=2Ω

Equivalent resistance between points C and B;

ohmslaw_2_solution2

 

 

 

Req=4+2=6Ω

b) V=I.R

36=I.6

I=6Amperes

3. a) Find equivalent resistance between the points X and Y.

b) If the current passing from 7Ω resistor is I1, and current passing from 8Ω resistor is I2, find I1/I2.

resistance_1

 

 

 

 

 

 

 

 

a) We redraw circuit and make it simple;

resistance_1_solution

 

 

 

 

 

 

 

 

 

Req=7+(2.2)/(2+2) (resistance in lower branch)

Req=8Ω

Equivalent resistance between points X and Y;

Req=(8.8)/(8+8)=4Ω

b)

resistance_1_solution2

 

 

 

 

 

Since resistances are in parallel, their potentials are equal. Using ohm's law;

V1=I1.R1 (lower branch)

V=I1.8

V2=I2.R2 (upper branch)

V=I2.8

I1/I2=1

4. If we close switch shown in the picture given below, find the changes in the brightness of the bulbs.

electriccircuit_1

 

 

 

 

 

 

 

 

Circuit when switch is opened;

electriccircuit_1_solution

 

 

 

 

 

 

 

 

Potential of B is equal to potential of D.

VA>VB=VD

When we close switch;

 electriccircuit_1_solution_2

 

 

 

 

 

 

potentials of the bulbs become;

VA=V

VB=VC=V/3

VD=2V/3

Thus; there is no change in the brightness of the bulb A, brightness of B decreases, and D increases.

5. Find the efficiency of the motor in the circuit given below.

resistance_2

 

 

 

 

 

 

 

 

 

Current passing through circuit is;

I=(ε1+ε2-ε')/Req

I=(40+60-70)/(7+2+2+4)

I=2 Amperes

Efficiency=ε'/(ε'+I.r')=70/(70+2.2)

Efficiency=35/37

 

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