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All questions of Unit 4: Electric Circuits for Grade 9 Exam
The current in a coil of resistance 90 ohms is to be reduced by 90 percent. What value of resistance should be connected in parallel with it?- a)100Ω
- b)10Ω
- c)9Ω
- d)90Ω
Correct answer is option 'B'. Can you explain this answer?
The current in a coil of resistance 90 ohms is to be reduced by 90 percent. What value of resistance should be connected in parallel with it?
a)
100Ω
b)
10Ω
c)
9Ω
d)
90Ω
|
Anjana Sharma answered |
The resistance is connected in parallel. Hence, voltage across the required resistance S and 90Ω is the same, i.e.
0.9 I x S = 0.1 I x 90
S = 90/9= 10Ω
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
Five resistances are connected as shown in fig. The effective resistance between points A and B is
- a)20/30Ω
- b)15Ω
- c)10/3Ω
- d)6Ω
Correct answer is option 'C'. Can you explain this answer?
Five resistances are connected as shown in fig. The effective resistance between points A and B is
a)
20/30Ω
b)
15Ω
c)
10/3Ω
d)
6Ω
|
Lekshmi Basu answered |
The given circuit is a balanced Wheatstone bridge type, hence it can be simplified 
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.
Three resistors of 4Ω, 12Ω , and 6Ω are connected in parallel. No. of 12Ω resistors required to be connected in parallel to reduce the total resistance to half of its original is- a)6
- b)3
- c)12
- d)2
Correct answer is 'A'. Can you explain this answer?
Three resistors of 4Ω, 12Ω , and 6Ω are connected in parallel. No. of 12Ω resistors required to be connected in parallel to reduce the total resistance to half of its original is
a)
6
b)
3
c)
12
d)
2
|
|
Riya Banerjee answered |
Here 4 Ω, 12 Ω, 6 Ω when connected in parallel results in 2Ω. to reduce it to half we have to join 1\R original = 6\12 for reducing it to half we have to join 6 , 12 Ω resistors in parallel (6\12) + (1\ 12 × 6) = 12\12 = 1 ohm . Half of its original value therefore option a is correct.
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.
Can you explain the answer of this question below:What is current I in the circuit as shown in figure?
- A:
1 A
- B:
2.0 A
- C:
1.2 A
- D:
0.5 A
The answer is b.
What is current I in the circuit as shown in figure?
1 A
2.0 A
1.2 A
0.5 A
|
Surya answered |
Yes option B is correct... first u straight the three resistance then it change in series... so u add three u get 6 ohms... after, that 6 ohm is parallel to 3ohm so;3/2.. so as per ohms law;V=IR (since:V=3; R=3/2; I=?) 3=3/2×I I=2ampere... that's it...hope u clear...!!👍😊
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.
In a meter bridge experiment a balance point is obtained at a distance of 60 cm from the left end when unknown resistance R is in a left gap and 8 ohms resistor is connected in the right gap. When the position of R and 8 ohm resistor is interchanged the balance point will be at distance of- a)40 cm
- b)30 cm
- c)50 cm
- d)60 cm
Correct answer is option 'A'. Can you explain this answer?
In a meter bridge experiment a balance point is obtained at a distance of 60 cm from the left end when unknown resistance R is in a left gap and 8 ohms resistor is connected in the right gap. When the position of R and 8 ohm resistor is interchanged the balance point will be at distance of
a)
40 cm
b)
30 cm
c)
50 cm
d)
60 cm
|
Jyoti Sengupta answered |
60/40 = l/100-l, = R/8
so R=12
now changed position 8 is proportional to l and 12 to (100-l)
8/12 = l/100-l
l = 40ohm
What is current I in the circuit as shown in figure?
- a)1 A
- b)2.0 A
- c)1.2 A
- d)0.5 A
Correct answer is option 'B'. Can you explain this answer?
What is current I in the circuit as shown in figure?
a)
1 A
b)
2.0 A
c)
1.2 A
d)
0.5 A
|
Preeti Iyer answered |
Three 2Ω resistors are in series. Their total resistance =6Ω. Now it is in parallel with 2Ω resistor, so total resistance,
1/R=1/2+1/6=3+1/6=4/6=2/3
R=3/2
∴I=RV=3/(3/2)=3×2/3=2A
1/R=1/2+1/6=3+1/6=4/6=2/3
R=3/2
∴I=RV=3/(3/2)=3×2/3=2A
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.
Meter Bridge is used to- a)determine unknown voltage v
- b)determine unknown resistance R
- c)determine unknown power P
- d)determine unknown emf e
Correct answer is option 'B'. Can you explain this answer?
Meter Bridge is used to
a)
determine unknown voltage v
b)
determine unknown resistance R
c)
determine unknown power P
d)
determine unknown emf e
|
Maulik Mehra answered |
Explanation:With a known resistance in one of the gaps, the meter bridge is used to determine the value of unknown resistance by the formula.
Three resistors of 4Ω, 12Ω , and 6Ω are connected in parallel. No. of 12Ω resistors required to be connected in parallel to reduce the total resistance to half of its original is- a)6
- b)3
- c)12
- d)2
Correct answer is 'A'. Can you explain this answer?
Three resistors of 4Ω, 12Ω , and 6Ω are connected in parallel. No. of 12Ω resistors required to be connected in parallel to reduce the total resistance to half of its original is
a)
6
b)
3
c)
12
d)
2
|
Sanchita Iyer answered |
Understanding Parallel Resistance
In a parallel circuit, the total resistance (R_total) is calculated using the formula:
1/R_total = 1/R1 + 1/R2 + 1/R3 + ...
For the given resistors:
- R1 = 4 ohms
- R2 = 12 ohms
- R3 = 6 ohms
Calculating the Original Total Resistance
Let's find the total resistance with the existing resistors.
- 1/R_total = 1/4 + 1/12 + 1/6
Finding a common denominator (which is 12):
- 1/R_total = 3/12 + 1/12 + 2/12 = 6/12
Thus,
- R_total = 2 ohms.
Target Resistance
We want to reduce the total resistance to half of its original value:
- Target Resistance = 2 ohms / 2 = 1 ohm.
Adding 12 Ohm Resistors
Let 'n' be the number of additional 12 ohm resistors needed in parallel to reach 1 ohm.
The new total resistance formula becomes:
1/R_new = 1/R_total + n/R2
Where R2 is 12 ohms:
1/R_new = 1/2 + n/12.
Setting R_new to 1 ohm gives:
1 = 1/2 + n/12.
Rearranging the equation:
1 - 1/2 = n/12
1/2 = n/12
Thus, n = 6.
Conclusion
To achieve a total resistance of 1 ohm, you need to connect 6 additional 12 ohm resistors in parallel. Hence, the correct answer is:
A) 6
In a parallel circuit, the total resistance (R_total) is calculated using the formula:
1/R_total = 1/R1 + 1/R2 + 1/R3 + ...
For the given resistors:
- R1 = 4 ohms
- R2 = 12 ohms
- R3 = 6 ohms
Calculating the Original Total Resistance
Let's find the total resistance with the existing resistors.
- 1/R_total = 1/4 + 1/12 + 1/6
Finding a common denominator (which is 12):
- 1/R_total = 3/12 + 1/12 + 2/12 = 6/12
Thus,
- R_total = 2 ohms.
Target Resistance
We want to reduce the total resistance to half of its original value:
- Target Resistance = 2 ohms / 2 = 1 ohm.
Adding 12 Ohm Resistors
Let 'n' be the number of additional 12 ohm resistors needed in parallel to reach 1 ohm.
The new total resistance formula becomes:
1/R_new = 1/R_total + n/R2
Where R2 is 12 ohms:
1/R_new = 1/2 + n/12.
Setting R_new to 1 ohm gives:
1 = 1/2 + n/12.
Rearranging the equation:
1 - 1/2 = n/12
1/2 = n/12
Thus, n = 6.
Conclusion
To achieve a total resistance of 1 ohm, you need to connect 6 additional 12 ohm resistors in parallel. Hence, the correct answer is:
A) 6
The Wheatstone bridge is balanced for four resistors R1,R2,R3 and R4 with a cell of emf 1.46 V. The cell is now replaced by another cell of emf 1.08 V. To obtain the balance again
- a)All the four resistance should be changed
- b)Both the resistance R1 and R4 should be changed
- c)No resistance needs to be changed
- d)Resistance R4 should be changed only
Correct answer is option 'C'. Can you explain this answer?
The Wheatstone bridge is balanced for four resistors R1,R2,R3 and R4 with a cell of emf 1.46 V. The cell is now replaced by another cell of emf 1.08 V. To obtain the balance again
a)
All the four resistance should be changed
b)
Both the resistance R1 and R4 should be changed
c)
No resistance needs to be changed
d)
Resistance R4 should be changed only
|
Neha Sharma answered |
The balance point of the Wheatstone’s bridge is determined by the ratio of the resistances. The change in the emf of the external battery will have no effect on the balance point.
Explanation:
- Initial Balanced Wheatstone Bridge: In the initial balanced Wheatstone bridge configuration, the emf of the cell is 1.46 V and all four resistors R1, R2, R3, and R4 are set to specific values to achieve balance.
- Replacement of Cell: When the cell is replaced by another cell with an emf of 1.08 V, the balance of the Wheatstone bridge is disrupted.
- Requirement for Rebalancing: In order to rebalance the Wheatstone bridge with the new cell of emf 1.08 V, no resistance needs to be changed.
- Reasoning: The balance of the Wheatstone bridge is determined by the ratio of the resistances in the bridge arms and not by the absolute values of the resistances. As long as the ratio of the resistances remains the same, the balance will be maintained regardless of the emf of the cell.
- Conclusion: Therefore, in this scenario, no resistance needs to be changed to obtain the balance again with the new cell of emf 1.08 V.
A voltmeter has a resistance of G ohm and range V volt. The value of resistance used in series to convert it into voltmeter of range nV volt is- a)nG
- b)(n−1)G
- c)G/n
- d)G/(n−1)
Correct answer is option 'B'. Can you explain this answer?
A voltmeter has a resistance of G ohm and range V volt. The value of resistance used in series to convert it into voltmeter of range nV volt is
a)
nG
b)
(n−1)G
c)
G/n
d)
G/(n−1)
|
|
Avantika Mehta answered |
Understanding Voltmeter Conversion
To convert a voltmeter of range V volts into a voltmeter of range nV volts, we need to consider the resistance of the voltmeter and the additional resistance required in series.
Principle of Operation
- A voltmeter measures the potential difference across its terminals.
- When converting to a higher range, we add a series resistance to ensure that the voltmeter can handle the increased voltage without damaging its internal components.
Given Parameters
- Resistance of the voltmeter = G ohm
- Original range of the voltmeter = V volts
- New desired range = nV volts
Calculating Series Resistance
1. Voltage Division: When a voltmeter with resistance G is connected in series with another resistance R, the voltage drop across the voltmeter is a fraction of the total voltage.
2. Using Voltage Ratios: The voltage across the voltmeter can be given by the ratio:
- V / (V + R) = G / (G + R)
3. Setting Up the Equation: For the new range of nV:
- nV / (nV + R) = G / (G + R)
4. Solving for R: Rearranging gives us:
- R = (n-1)G
This means the resistance required in series to achieve the desired range of nV volts is (n - 1)G ohms.
Conclusion
Thus, the correct answer for the resistance used in series to convert the voltmeter into one of range nV volts is option B: (n - 1)G. This ensures proper functioning and safety of the voltmeter under higher voltage conditions.
To convert a voltmeter of range V volts into a voltmeter of range nV volts, we need to consider the resistance of the voltmeter and the additional resistance required in series.
Principle of Operation
- A voltmeter measures the potential difference across its terminals.
- When converting to a higher range, we add a series resistance to ensure that the voltmeter can handle the increased voltage without damaging its internal components.
Given Parameters
- Resistance of the voltmeter = G ohm
- Original range of the voltmeter = V volts
- New desired range = nV volts
Calculating Series Resistance
1. Voltage Division: When a voltmeter with resistance G is connected in series with another resistance R, the voltage drop across the voltmeter is a fraction of the total voltage.
2. Using Voltage Ratios: The voltage across the voltmeter can be given by the ratio:
- V / (V + R) = G / (G + R)
3. Setting Up the Equation: For the new range of nV:
- nV / (nV + R) = G / (G + R)
4. Solving for R: Rearranging gives us:
- R = (n-1)G
This means the resistance required in series to achieve the desired range of nV volts is (n - 1)G ohms.
Conclusion
Thus, the correct answer for the resistance used in series to convert the voltmeter into one of range nV volts is option B: (n - 1)G. This ensures proper functioning and safety of the voltmeter under higher voltage conditions.
A piece of copper and another of germanium are cooled from room temperature to 80K. The resistance- a)of each of them decreases
- b)of each of them increases
- c)copper increases and that of germanium decreases
- d)copper decreases and that of germanium increases
Correct answer is option 'D'. Can you explain this answer?
A piece of copper and another of germanium are cooled from room temperature to 80K. The resistance
a)
of each of them decreases
b)
of each of them increases
c)
copper increases and that of germanium decreases
d)
copper decreases and that of germanium increases
|
Pranjal Pillai answered |
Explanation:Copper is a conductor and we know that for conductors, resistance is directly proprtional to temperature. Therefore on decreasing temperature resistance also decreases.Whereas, germanium is a semiconductor and for semiconductors, resistance is inversely proportional to temperature. So on decreasing temperature resistance increases.
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
In a metre bridge experiment, null point is obtained at 20 cm from one end of the wire when resistance X is balanced against another resistance Y. If X<Y, then where will be the new position of the null point from the same end, if one decides to balance a resistance of 4X against Y ?- a)50 cm
- b)80 cm
- c)40 cm
- d)70 cm
Correct answer is option 'A'. Can you explain this answer?
In a metre bridge experiment, null point is obtained at 20 cm from one end of the wire when resistance X is balanced against another resistance Y. If X<Y, then where will be the new position of the null point from the same end, if one decides to balance a resistance of 4X against Y ?
a)
50 cm
b)
80 cm
c)
40 cm
d)
70 cm
|
|
Pranab Kapoor answered |
Understanding the Metre Bridge Experiment
In a metre bridge experiment, the balance point (null point) indicates the equality of the ratios of resistances. Here’s a detailed explanation of how to determine the new null point when balancing different resistances.
Initial Setup
- A null point is found at 20 cm from one end of the wire when resistance X is balanced against resistance Y.
- This can be expressed as:
(X / Y) = (Length from one end / Length from the other end)
Thus, (X / Y) = (20 cm / 80 cm) = 1/4.
- Given that X < y,="" we="" can="" infer="" that="" x="" />
New Resistance Balancing
- Now, we are tasked with balancing a resistance of 4X against Y.
- The new equation becomes (4X / Y) = (Length from one end / Length from the other end).
Calculating New Null Point
- From the previous relationship, we know that X = (1/4)Y, therefore:
4X = 4 * (1/4)Y = Y.
- This implies that when balancing 4X against Y, they are equal.
- Hence, the new null point would be halfway along the bridge length, which is at 50 cm from one end.
Conclusion
- Therefore, when balancing a resistance of 4X against Y, the new null point will be at 50 cm from the original end.
- Thus, the correct answer is option a) 50 cm.
In a metre bridge experiment, the balance point (null point) indicates the equality of the ratios of resistances. Here’s a detailed explanation of how to determine the new null point when balancing different resistances.
Initial Setup
- A null point is found at 20 cm from one end of the wire when resistance X is balanced against resistance Y.
- This can be expressed as:
(X / Y) = (Length from one end / Length from the other end)
Thus, (X / Y) = (20 cm / 80 cm) = 1/4.
- Given that X < y,="" we="" can="" infer="" that="" x="" />
New Resistance Balancing
- Now, we are tasked with balancing a resistance of 4X against Y.
- The new equation becomes (4X / Y) = (Length from one end / Length from the other end).
Calculating New Null Point
- From the previous relationship, we know that X = (1/4)Y, therefore:
4X = 4 * (1/4)Y = Y.
- This implies that when balancing 4X against Y, they are equal.
- Hence, the new null point would be halfway along the bridge length, which is at 50 cm from one end.
Conclusion
- Therefore, when balancing a resistance of 4X against Y, the new null point will be at 50 cm from the original end.
- Thus, the correct answer is option a) 50 cm.
According to Kirchhoff’s Loop Rule
- a)The absolute sum of changes in potential around any closed loop must be same.
- b)The algebraic sum of changes in potential around any closed loop must be negative.
- c)The algebraic sum of changes in potential around any closed loop must be positive.
- d)The algebraic sum of changes in potential around any closed loop must be zero.
Correct answer is option 'D'. Can you explain this answer?
According to Kirchhoff’s Loop Rule
a)
The absolute sum of changes in potential around any closed loop must be same.
b)
The algebraic sum of changes in potential around any closed loop must be negative.
c)
The algebraic sum of changes in potential around any closed loop must be positive.
d)
The algebraic sum of changes in potential around any closed loop must be zero.
|
|
Neha Chauhan answered |
Kirchhoff’s loop rule is based on the principle of conservation of energy. The work done in transporting a charge in a closed loop is zero. The algebraic sum ( since potential differences can be both positive and negative) of potential differences around any closed loop is always zero.
Potentiometer measures the potential difference more accurately than a voltmeter, because- a)It draws a heavy current from external circuit.
- b)It does not draw current from external circuit.
- c)it has a wire of low resistance.
- d)it has a wire of high resistance
Correct answer is option 'B'. Can you explain this answer?
Potentiometer measures the potential difference more accurately than a voltmeter, because
a)
It draws a heavy current from external circuit.
b)
It does not draw current from external circuit.
c)
it has a wire of low resistance.
d)
it has a wire of high resistance
|
Amar Pillai answered |
Explanation:Potentiometer measures the potential difference using null deflection method, where no current is drawn from the cell; whereas voltmeter needs a small current to show deflection. So, accurate measurement of p.d is done using a potentiometer.
Drift is the random motion of the charged particles within a conductor,- a)along with a very slow net motion in the opposite direction of the field
- b)along with zero motion in the direction of the field
- c)along with a decelerated motion in the direction of the field
- d)along with accelerated motion in the direction of the field
Correct answer is option 'A'. Can you explain this answer?
Drift is the random motion of the charged particles within a conductor,
a)
along with a very slow net motion in the opposite direction of the field
b)
along with zero motion in the direction of the field
c)
along with a decelerated motion in the direction of the field
d)
along with accelerated motion in the direction of the field
|
|
Pranjal Pillai answered |
Explanation:The electrons in a conductor have random velocities and when an electric field is applied, they suffer repeated collisions and in the process move with a small average velocity, opposite to the direction of the field. This is equivalent to positive charge flowing in the direction of the field.
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.
In a potentiometer experiment, for measuring internal resistance of a cell, the balance point has been obtained on the fourth wire. The balance point can be shifted to fifth wire by- a)increasing the current due to auxiliary battery
- b)Putting a shunt resistance in parallel with the cell.
- c)decreasing the current due to auxiliary battery
- d)putting a suitable resistance in series with the cell.
Correct answer is option 'C'. Can you explain this answer?
In a potentiometer experiment, for measuring internal resistance of a cell, the balance point has been obtained on the fourth wire. The balance point can be shifted to fifth wire by
a)
increasing the current due to auxiliary battery
b)
Putting a shunt resistance in parallel with the cell.
c)
decreasing the current due to auxiliary battery
d)
putting a suitable resistance in series with the cell.
|
|
Pranjal Pillai answered |
Explanation:If the current due to the auxillary battery is decreased, the potential gradient will be decreases, so the balancing length increases.Thus null point will move to fifth wire.
The sensitivity of the potentiometer can be increased by:- a)increasing the length of potentiometer wire.
- b)increasing the e.m.f. of primary cell.
- c)decreasing the length of potentiometer wire.
- d)increasing the potential gradient.
Correct answer is option 'A'. Can you explain this answer?
The sensitivity of the potentiometer can be increased by:
a)
increasing the length of potentiometer wire.
b)
increasing the e.m.f. of primary cell.
c)
decreasing the length of potentiometer wire.
d)
increasing the potential gradient.
|
Anjali Reddy answered |
Explanation:A potentiometer is considered to be sensitive if the potential gradient dV/dl is low. Such a potentiometer can measure very small changes in potential difference. Increasing the length of the potentiometer wire decreases the potential gradient. Its sensitivity increases. Increasing potential gradient decreases the sensitivity. increasing the emf of the primary cell and by decreasing the length, potential gradient increases.
In the figure, voltmeter and ammeter shown are ideal. Then voltmeter and ammeter readings, respectively, are
- a)125 V,3 A
- b)100 V,4 A
- c)120 V,4 A
- d)120 V,3 A
Correct answer is option 'B'. Can you explain this answer?
In the figure, voltmeter and ammeter shown are ideal. Then voltmeter and ammeter readings, respectively, are
a)
125 V,3 A
b)
100 V,4 A
c)
120 V,4 A
d)
120 V,3 A
|
Ambition Institute answered |
Resistors 20Ω,100Ω and 25Ω will be in parallel. Their equivalent is 10Ω.
p.d. across 10Ω,10I = 10 × 10 = 100 V
This will be the voltmeter reading. Also, this will be the p.d. across each of 20Ω,100Ω and 25Ω resistors.
Ammeter reading = current through 25Ω=100/25=4 A.
p.d. across 10Ω,10I = 10 × 10 = 100 V
This will be the voltmeter reading. Also, this will be the p.d. across each of 20Ω,100Ω and 25Ω resistors.
Ammeter reading = current through 25Ω=100/25=4 A.
Three identical cells, each having an e.m.f. of 1.5 V and a constant internal resistance of 2.0Ω, are connected in series with a 4.0Ω resistor R, first as in circuit (i), and secondly as in circuit (ii).
What is the ratio 
- a)9.0
- b)7.2
- c)1.8
- d)3.0
Correct answer is option 'A'. Can you explain this answer?
Three identical cells, each having an e.m.f. of 1.5 V and a constant internal resistance of 2.0Ω, are connected in series with a 4.0Ω resistor R, first as in circuit (i), and secondly as in circuit (ii).
What is the ratio
What is the ratio
a)
9.0
b)
7.2
c)
1.8
d)
3.0
|
Mohit Rajpoot answered |
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