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All questions of Unit 3: Electric Circuits for Grade 9 Exam
V-I graph of which material shows the straight line- a)Silicon
- b)Silver
- c)Germanium
- d)Gallium
Correct answer is option 'B'. Can you explain this answer?
V-I graph of which material shows the straight line
a)
Silicon
b)
Silver
c)
Germanium
d)
Gallium
|
Riya Banerjee answered |
Materials which obeys the ohm’s law show straight line in the V-I graph. Since silver is the only ohmic material in given options, it shows straight line curve.
Which of the following is an ohmic conductor?- a)LED
- b)Thyristor
- c)diode
- d)Metal conductor
Correct answer is option 'D'. Can you explain this answer?
Which of the following is an ohmic conductor?
a)
LED
b)
Thyristor
c)
diode
d)
Metal conductor
|
Geetika Shah answered |
Ohmic conductors are those which follows ohm's Law
Constantan is an copper nickle alloy which follows ohm's law
An electrolyte is a chemical that produces an electrically conducting solution and hence conducts electrical current.
But Electrolyte can be Ohmic as well as non-ohmic conductor
Transistor is semi-conductor device and does not follow ohm's Law
thermionic valves - is vacuum tube (electronic tube) , uses ion emission.
So option C Constanton s an ohmic conductor.
The following fig. shows I-V graph for a given metallic wire at two temperatures T1and T2.Then,
- a)Temperature T2 is less
- b)Temperature T2 is more
- c)T1 is same as T2
- d)None of the above
Correct answer is option 'B'. Can you explain this answer?
The following fig. shows I-V graph for a given metallic wire at two temperatures T1and T2.Then,
a)
Temperature T2 is less
b)
Temperature T2 is more
c)
T1 is same as T2
d)
None of the above
|
Knowledge Hub answered |
The slope of the graph = I / V
In case of T2, I/V is less. Hence, V/I (=R) is more.
Value of resistance increases with increasing temperature.
Hence, T2 >T
How many electrons are flowing per second through a section of a conductor corresponding to current of 1A?- a)7. 00 × 1018
- b)6.25 × 1018
- c)3.25 × 1016
- d)4.75 × 1017
Correct answer is 'B'. Can you explain this answer?
How many electrons are flowing per second through a section of a conductor corresponding to current of 1A?
a)
7. 00 × 1018
b)
6.25 × 1018
c)
3.25 × 1016
d)
4.75 × 1017
|
Vishal Giri answered |
I = q / t but we know q = n*e# where n = no. of electrons and e = charge on one electron. Let's put:: q=n*e we get I = n*e / t we know from given information that =============>I = 1A,,,,, t =1 s,,,, e = 1.602 * 10^ -19 C. put all these values in above equation we get 1 = n* 1.6 *10^ -19/ 1 hence n = 1 / 1.6*10^ -19 n = 0.625 * 10^ 19 n = 6.25 * 10^ 18IS THE UR ANSWER.
Two cells of 1.25 V and 0.75V are connected in parallel. The effective voltage will be- a)0.5 V
- b)2.00 V
- c)1.25 V
- d)0.75 V
Correct answer is option 'A'. Can you explain this answer?
Two cells of 1.25 V and 0.75V are connected in parallel. The effective voltage will be
a)
0.5 V
b)
2.00 V
c)
1.25 V
d)
0.75 V
|
Anjana Sharma answered |
When two cells are conneted in parallel , then
Effective Voltage,V = V1 − V2 = 1.25 − 0.75 = 0.5 V
1 ohm is equal to- a)1 volt per ampere
- b)1 ampere per millivolt
- c)1 milliampere per volt
- d)1 ampere per volt
Correct answer is option 'A'. Can you explain this answer?
1 ohm is equal to
a)
1 volt per ampere
b)
1 ampere per millivolt
c)
1 milliampere per volt
d)
1 ampere per volt
|
|
Anjana Sharma answered |
Reduced to base SI units, one ohm is the equivalent of one kilogram meter squared per second cubed per ampere squared (1 kg times m. s. A^-2 . The ohm is also the equivalent of a volt per ampere (V/A).
At any junction, the sum of the currents entering the junction is equal to the sum of _______- a)potential around any closed loop
- b)voltages across the junction
- c)all the currents in the circuit
- d)currents leaving the junction
Correct answer is option 'D'. Can you explain this answer?
At any junction, the sum of the currents entering the junction is equal to the sum of _______
a)
potential around any closed loop
b)
voltages across the junction
c)
all the currents in the circuit
d)
currents leaving the junction
|
Om Kumar answered |
The correct answer is option 'D': currents leaving the junction.
Explanation:
At any junction in an electrical circuit, the sum of the currents entering the junction is equal to the sum of the currents leaving the junction. This is based on the principle of conservation of charge.
When current flows through a junction, it must split into multiple paths. The total amount of charge entering the junction must be equal to the total amount of charge leaving the junction. This is because charge cannot be created or destroyed, it can only flow through the circuit.
To better understand this concept, consider a simple circuit with three branches connected to a junction. Let's label the currents entering the junction as I1, I2, and I3, and the currents leaving the junction as I4, I5, and I6.
The principle of conservation of charge states that the total amount of charge entering the junction must be equal to the total amount of charge leaving the junction. Mathematically, this can be expressed as:
I1 + I2 + I3 = I4 + I5 + I6
This equation shows that the sum of the currents entering the junction (I1 + I2 + I3) is equal to the sum of the currents leaving the junction (I4 + I5 + I6).
This principle is a consequence of Kirchhoff's current law (KCL), which states that the algebraic sum of currents at any junction in an electrical circuit is zero. This means that the sum of currents entering the junction is equal to the sum of currents leaving the junction.
In summary, at any junction in an electrical circuit, the sum of the currents entering the junction is equal to the sum of the currents leaving the junction. This principle is based on the conservation of charge and is a consequence of Kirchhoff's current law.
Explanation:
At any junction in an electrical circuit, the sum of the currents entering the junction is equal to the sum of the currents leaving the junction. This is based on the principle of conservation of charge.
When current flows through a junction, it must split into multiple paths. The total amount of charge entering the junction must be equal to the total amount of charge leaving the junction. This is because charge cannot be created or destroyed, it can only flow through the circuit.
To better understand this concept, consider a simple circuit with three branches connected to a junction. Let's label the currents entering the junction as I1, I2, and I3, and the currents leaving the junction as I4, I5, and I6.
The principle of conservation of charge states that the total amount of charge entering the junction must be equal to the total amount of charge leaving the junction. Mathematically, this can be expressed as:
I1 + I2 + I3 = I4 + I5 + I6
This equation shows that the sum of the currents entering the junction (I1 + I2 + I3) is equal to the sum of the currents leaving the junction (I4 + I5 + I6).
This principle is a consequence of Kirchhoff's current law (KCL), which states that the algebraic sum of currents at any junction in an electrical circuit is zero. This means that the sum of currents entering the junction is equal to the sum of currents leaving the junction.
In summary, at any junction in an electrical circuit, the sum of the currents entering the junction is equal to the sum of the currents leaving the junction. This principle is based on the conservation of charge and is a consequence of Kirchhoff's current law.
The Wheatstone bridge Principle is deduced using- a)Gauss’s Law
- b)Kirchhoff’s Laws
- c)Coulomb’s Law
- d)Newton’s Laws
Correct answer is option 'B'. Can you explain this answer?
The Wheatstone bridge Principle is deduced using
a)
Gauss’s Law
b)
Kirchhoff’s Laws
c)
Coulomb’s Law
d)
Newton’s Laws
|
|
Anjana Sharma answered |
PRINCIPLE: Wheatstone bridge principle states that when the bridge is balanced, the product of the resistance of the opposite arms are equal. The files that I had attached in which I had derived Wheatstone bridge equation using Kirchhoff law is useful to you.
Can you explain the answer of this question below:The ______ of changes in potential around any closed loop involving resistors and cells in a loop is zero.- A:product
- B:algebraic sum
- C:difference
- D:sum of absolute values
The answer is b.
The ______ of changes in potential around any closed loop involving resistors and cells in a loop is zero.
A:
product
B:
algebraic sum
C:
difference
D:
sum of absolute values
|
Lavanya Menon answered |
In accordance with Kirchhoff’s second law i.e. Kirchhoff’s voltage law (KVL), the algebraic sum of all the potential differences in a closed electric circuit or closed loop that contains one or more cells and resistors is always equal to zero.
This law is popularly called the law of conservation of voltage.
This law is popularly called the law of conservation of voltage.
The following fig. shows I-V graph for a given metallic wire at two temperatures T1and T2.Then,
- a)Temperature T2 is less
- b)Temperature T2 is more
- c)T1 is same as T2
- d)None of the above
Correct answer is option 'C'. Can you explain this answer?
The following fig. shows I-V graph for a given metallic wire at two temperatures T1and T2.Then,
a)
Temperature T2 is less
b)
Temperature T2 is more
c)
T1 is same as T2
d)
None of the above
|
|
Rahul Bansal answered |
The slope of the given graph gives us the inverse of resistance. Resistance of a material increases with increasing temperature because the collision between the molecules increases.
In the graph given, T2 has a smaller slope and hence corresponds to higher resistance. Therefore, T2 > T1.
Given N resistors each of resistance R are first combined to get minimum possible resistance and then combined to get maximum possible resistance. The ratio of the minimum to maximum resistance is- a)N
- b)N2
- c)1/N2
- d)1/N
Correct answer is option 'C'. Can you explain this answer?
Given N resistors each of resistance R are first combined to get minimum possible resistance and then combined to get maximum possible resistance. The ratio of the minimum to maximum resistance is
a)
N
b)
N2
c)
1/N2
d)
1/N
|
Nishtha Bose answered |
They are connected in series to get maximum in this case resistance would be nr
and to get minimun resistance they are connected in parallel :. resistance in this case is n/r
:. ratio between minimum and maximum resistance is n/r/nr = 1/r^2
Can you explain the answer of this question below:The following fig. shows I-V graph for a given metallic wire at two temperatures T1and T2.Then,
- A:
Temperature T2 is less
- B:
Temperature T2 is more
- C:
T1 is same as T2
- D:
None of the above
The answer is c.
The following fig. shows I-V graph for a given metallic wire at two temperatures T1and T2.Then,
Temperature T2 is less
Temperature T2 is more
T1 is same as T2
None of the above
|
Juhi Deshpande answered |
Temperature T₁ is higher than the Temperature T₂, this is because If the temperature will increase then by the concept of the the Heating effect of the current, there will be rise in current. Thus, the graph will be having the more slope.
For the Temperature T₂, line is having a small slope than the temperature T₁ because of the high temp., heating is produced as a result current has been increased.
Note ⇒ Ohm's law is only valid in case, the temperature remains constant, but in this expression temperature is not constant therefore, ohm's law will be not valid. But it can me make valid on different temperatures at a time.
Chapter doubts & questions for Unit 3: Electric Circuits - AP Physics C Electricity and Magnetism 2025 is part of Grade 9 exam preparation. The chapters have been prepared according to the Grade 9 exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for Grade 9 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.
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AP Physics C Electricity and Magnetism
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