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Yogesh answered on Apr 21 2021
4-8-question-of-the-day.docx
Ques. How do you find equivalent resistance in a series circuit?
Solution:
The resistance is said to be in series when the terminal of one resistance is connected with other terminal. And current (I) flowing through all resistances is same.
If we consider three resistances R1, R2, R3 are connected in series with a power supply of V volts. Then by Ohm’s law we know that voltage is directly proportional to current.
V=I*R
Then voltage through R1, is
V1 = I*R1
Also, V2 = I* R2
V3 = I*R3
Therefore, V= V1+V2+V3
V= I*R1+I*R2+I*R3
V=I (R1+R2+R3) …………(1)
Consider equivalent resistance is Rs,
Therefore, from eq. (1)
I*Rs= I (R1+R2+R3)
Rs =R1+R2+R3 is the equivalent resistance.
4-16-question-of-the-day.docx
Que. What do you know about magnets?
· Magnets are the elements which attract materials like iron, nickel and cobalt.
· The material which shows the property of ferromagnetism is used to make permanent magnets.
· When a conductor carries a free electron it creates few magnetic lines of forces which are able to attract iron dust particles.
· There are two types of magnets temporary and permanent.
· Magnets have two poles South Pole and North Pole in both permanent and temporary magnets.
· Earth also has magnetism with the help of which compass is operated and it shows south and north direction.
· Permanent magnets are found in nature whereas temporary magnets are made artificially.
chapter-22-23-homework.docx
CHAPTER 22:
Que. 42- Ans:
Solution:
Electric current is measured in ampere whereas charge and time is measured in coulomb and second respectively.
Electric current is directly proportional to charge and inversely proportional to time. If one coulomb of charge is created for one second then one ampere of current will flow.
Que. 43- Ans:
Solution:
The voltmeter should be connected across or parallel to motor to measure voltage.
Que. 44- Ans:
Solution:
The ammeter should be connected in series with the circuit to measure the current.
Que. 45- Ans:
Solution:
The direction of current is from positive terminal of battery to negative terminal of battery through the motor.
Que. 46- Ans:
Solution:
a. Motor transforms electrical energy into mechanical energy.
b. Battery converts chemical energy into electrical energy.
c. Switch turns the circuit on and off.
d. Potentiometer is used to adjust speed.
Que. 47- Ans:
Solution:
We know that
R=r*l/A
Where R= Resistance, r= Resistivity, l= Length of wire, A= Cross section area of wire.
The wire with large cross section diameter will have least value of resistance because the cross section area is inversely proportional to resistance.
Que. 48- Ans:
Solution:
a.
b. The ammeter must be connected in series.
c. The voltmeter must be connected in parallel.
Que. 49- Ans:
Solution:
a. The potentiometer is variable resistance and hence by varying the value resistance the electric motor can be controlled.
b. The potentiometer is used to convert the motion from joystick to the TV screen.
Que. 50- Ans:
Solution:
a. Electrical energy is transformed into heat and light.
b. Electrical energy is transformed into heat.
c. Electrical energy to light and sound.
d. Chemical energy to light and thermal energy.
Que. 51- Ans:
Given Data: V = 12 V, I = 1.5 A
Find: a) Power delivered to the motor=P=?
b) Energy Transferred=? If t= 15 min=900 sec.
Solution:
a. We know that
Power = Voltage * Current
P=V*I
P=12*1.5
P=18 W
b. Energy = E = Power* Time
E=P*t
E= 18* 900
E=16,200 J.
Que. 52- Ans:
Given Data: V= 27V, R= 18 ohm
Find: a. Ammeter reading (I)
b. Voltmeter reading (V)
c. Power delivered to resistor (P)
d. Energy delivered to resistor per hour (E)
Solution:
a. We know by ohms law
V=I*R
I=V/R
I=27/18
I= 1.5A (Ammeter reading)
b. V=27 v (Voltmeter reading)
c. P= V*I
P=27*1.5
P=40.5 W (power delivered to resistor)
d. E= P*t
E= 40.5*3600 (since 1 hour=60 min=3600sec)
E= 145,800 J (Energy delivered to resistor per hour)
Que. 53- Ans:
Given Data: V =120 V
I =1A
Find: Power dissipated by toaster =?
Solution:
P = IV
P = 8.0*120 = 9.6×102 W
Que. 54- Ans:
Given Data: I = 1A
V= 120V
Find: Rate at which bulb transforms energy =?
Solution:
P = IV
P = 1.2 *120 = 1.4×102 W
Que. 55- Ans:
Given Data: V= 27V
R=9 ohm
Find: a. Ammeter reading (I)
b. Voltmeter reading(V)
c. Power delivered to resistor (P)
d. Energy delivered to resistor per hour(E)
Solution:
a. By ohm’s law
I=V/R
I= 27/9
I= 3A
b. V = 27 V
c. P = VI = 27 *3.0 = 81 W
d. E = P*t = 81*3600 = 2.9×105J
Que. 56- Ans:
Given Data: V= 9.0V, R=18 ohm
Find: a. Ammeter reading(I)
b. Voltmeter reading(V)
c. Power delivered to resistor(P)
d. Energy delivered to resistor per hour(E)
Solution:
a. By ohm’s law,
I=V/R
I=9/18
I= 0.5A
b. V = 9 V
c. P = VI = 9.0 *0.5 = 4.5 W
d. E = P*t = 4.5*3600 = 1.6×104 J
Que. 57- Ans:
Given Data: I=0.50A, V= 120V
Find: a. Power (P)
b. Energy transformed in 5 min(E)
Solution:
a. P = IV = 0.50*120 = 60 W
b. E= P* t
E= 60*300
E= 18000 J
Que. 58- Ans:
Given Data: I= 210A
V=12 V
Find: a. Energy per second=?
b. Power (P) =?
Solution:
Here, Power=Energy
Since t=1sec
a. P=E = IV = 210 *12 = 2500 J/s
b. P = 2500 W
Que. 59- Ans:
Given Data: P= 4200 W
V= 220 V
Find: I=?
Solution:
We know,
P= I*V
I=4200/220
I=19 A
Que. 60- Ans:
Given Data: I=1.5A, V= 3V
Find: a. Power (P) =?
b. Energy transform in 11 min=?
Solution:
a. P = IV = 1.5* 3.0 = 4.5 W
b. E= P* t
E= 4.5*11*60
E= 2970 J
Que. 61- Ans:
Given Data: R=60.0 ohm, I= 0.40A
Find: V=?
Solution:
We know by ohm’s Law
V = IR = 0.40 *60 = 24 V
Que. 62- Ans:
Given Data: R=60.0 ohm, I= 0.40A
Find: V=?
Solution:
We know by ohm’s Law
V = IR = 1.5*4 = 6 V
Que. 63- Ans:
Given Data: R=15 ohm, I= 8.0A
Find: V=?
Solution:
We know by ohm’s Law
V = IR = 8*15 = 120 V
Que. 64- Ans:
Given Data: R=150 ohm, V= 75V
Find: I=?
Solution:
I=V/R
I=75/150
I=0.5A
Que. 65- Ans:
Given Data: Table
Find: a. For each measurement resistance=?
b. Graph I versus
Solution:
a. R = 143 ohm, R = 148 ohm, R= 150 ohm, R = 154 ohm, R = 159 ohm, R = 143 ohm, R = 143 ohm, R = 154 ohm, R= 157 ohm, R = 161 ohm
b. Graph
c. No, it does not obey ohm’s law.
Que. 66- Ans:
Given Data: R=16 ohm, I= 1.75A
Solution:
V = IR = 1.75* 16 = 28 V
Que. 67- Ans:
Given Data: I= 66mA when V=6V
I=75mA when V=9V
Find: a. Does the lamp obey ohm’s law
b. Power (P) when lamp is connected to 6V battery
Solution:
a. No, the lamp does not obey ohm’s law
b. P = IV = 66×10−3* 6.0 = 0.40 W
Que. 68- Ans:
Given Data: P= 60 W
T= 30 min=1800sec
Bulb transforms its 12% electrical energy to radiant energy.
Find: Thermal Energy during Half an hour
Solution:
E=P*t
E=60*1800
E=108*103 J
If the bulb is 12% efficient, 88 % of the energy is lost in the form heat.
Therefore, Thermal Energy is
Q = 0.88*108×103 = 95×103 J
Que. 69- Ans:
Given Data: I= 0.40A when V=120V
Find: a. R=? When lamp is ON
b. When the lamp is cold, its resistance is 1/5 as great as it is when the lamp is hot. What is the lamp’s cold resistance?
c. What is the current through the lamp as it is turned on if it is connected to a potential difference of 120 V?
Solution:
a. V = IR
R=V/I
R= 120/0.40
R= 300 ohm
b. 1/5 * 300 = 60 ohm
c. I=V/R
I=120/60
I= 2A
Que. 70- Ans:
Given Data: A graph of diode
Find: a. R=? When V=0.7V
b. R=? When V= 0.6V
Solution:
a. From the graph,
I = 22 mA, and V = IR, so
R=V/I
R=0.7/22*10-3
R= 32 ohm
b. From the graph,
I = 5 mA, and V = IR, so
R=V/I
R=0.6/5*10-3
R= 120 ohm
Que. 71- Ans-
Given Data: V=90V, R= 45 ohm
Find: Ammeter reading (I)
Solution:
By ohms law,
I=V/R
I=90/45
I=2A
Que. 72- Ans:
Solution:
Due to low resistance of incandescent filament a sudden high current is passed through it which results to glow the light with a very temperature, this is the reason they burn out more frequently.
Que. 73- Ans:
Solution:
The resistance of Copper is very low so when both terminal of copper is connected to the battery it results in high current and hence due to this current a rapid heat is produced in copper.
Que. 74- Ans:
Solution:
The resistance of the wire must be kept small to transmit electrical energy economically over long distance.
Que. 75- Ans:
Solution:
Power is defined as energy per unit time
P= E/t
And energy is ½ MV2 , where M= mass and V=velocity
Also velocity is meter per second.
Therefore, P=J/s => MV2 /s => M*(m/s)2 /s ...