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When the position of cell and galvanometer in a Wheatstone bridge is inter-changed, its balanced condition- a)Changes and depends on galvanometer position only
- b)Changes
- c)Changes and it depends on cell position only
- d)Remains unchanged
Correct answer is option 'D'. Can you explain this answer?
When the position of cell and galvanometer in a Wheatstone bridge is inter-changed, its balanced condition
a)
Changes and depends on galvanometer position only
b)
Changes
c)
Changes and it depends on cell position only
d)
Remains unchanged
|
Krishna Iyer answered |
For balanced Wheatstone bridge which is shown in figure a, P/Q=S/R
If we interchange the cell and galvanometer then circuit becomes as shown in figure b.
and balanced condition, P/S=Q/R⇒P/Q=S/R
Thus, balanced point remains unchanged.
If we interchange the cell and galvanometer then circuit becomes as shown in figure b.
and balanced condition, P/S=Q/R⇒P/Q=S/R
Thus, balanced point remains unchanged.
When Wheatstone bridge is in balance condition, the current through galvanometer will be- a)Zero
- b)Maximum
- c)Minimum
- d)Depends upon the type of galvanometer
Correct answer is option 'A'. Can you explain this answer?
When Wheatstone bridge is in balance condition, the current through galvanometer will be
a)
Zero
b)
Maximum
c)
Minimum
d)
Depends upon the type of galvanometer
|
|
Krishna Iyer answered |
The bridge is said to be balanced when deflection in galvanometer is zero(Ig=0), i.e., no current flows through the galvanometer(branch BD).In the balanced condition,
P/Q=R/S
On mutually changing the position of the cell and galvanometer, this condition will not change.
Meter Bridge or Slide Wire Bridge is a practical form of- a)Ammeter
- b)Voltmeter
- c)Wheatstone bridge
- d)Potentiometer
Correct answer is option 'C'. Can you explain this answer?
Meter Bridge or Slide Wire Bridge is a practical form of
a)
Ammeter
b)
Voltmeter
c)
Wheatstone bridge
d)
Potentiometer
|
Aarav Khanna answered |
Meter Bridge or Slide Wire Bridge is a practical form of Wheatstone bridge.
Wheatstone Bridge:
The Wheatstone bridge is a circuit used to measure unknown resistance by balancing it against a known resistance. It consists of four resistors arranged in a diamond shape, with a galvanometer connected between two opposite corners and a battery connected to the other two corners. When the bridge is balanced, the galvanometer shows zero deflection, indicating that the ratio of the unknown resistance to the known resistance is equal to the ratio of the other two resistors.
Meter Bridge or Slide Wire Bridge:
The meter bridge, also known as the slide wire bridge, is a practical form of the Wheatstone bridge. It consists of a long uniform wire of uniform cross-section called the meter wire, which is stretched over a wooden board. The meter wire is usually made of manganin or constantan, which have low temperature coefficients of resistance.
Construction:
The meter bridge consists of a uniform wire AB, which is connected to a galvanometer G at its midpoint. The wire is divided into two parts by a gap at the center, where a resistance box is connected. A jockey J is used to make contact with the wire and slide along its length.
Working Principle:
To measure an unknown resistance using the meter bridge, the jockey is initially placed at the midpoint of the wire (point O). The resistance box is adjusted until the galvanometer shows zero deflection. At this point, the bridge is in a balanced condition.
The principle behind the working of the meter bridge is that when the bridge is balanced, the ratio of the lengths of the two arms of the wire (AO and OB) is equal to the ratio of the resistances (R1 and R2) in the two arms. Mathematically, this can be represented as:
R1/R2 = AO/OB
By measuring the lengths AO and OB, the ratio R1/R2 can be determined. Since one of the resistances (R1 or R2) is known, the unknown resistance can be calculated.
Advantages of Meter Bridge:
1. High Accuracy: The meter bridge offers high accuracy in measuring resistance as it is based on the principle of a balanced bridge.
2. Simple Construction: The meter bridge is relatively simple in construction and easy to use.
3. Low Cost: The materials required for constructing a meter bridge are easily available and inexpensive.
4. Versatility: The meter bridge can be used to measure a wide range of resistances, making it a versatile instrument.
In conclusion, the meter bridge or slide wire bridge is a practical form of the Wheatstone bridge and is used to measure unknown resistances accurately.
Wheatstone Bridge:
The Wheatstone bridge is a circuit used to measure unknown resistance by balancing it against a known resistance. It consists of four resistors arranged in a diamond shape, with a galvanometer connected between two opposite corners and a battery connected to the other two corners. When the bridge is balanced, the galvanometer shows zero deflection, indicating that the ratio of the unknown resistance to the known resistance is equal to the ratio of the other two resistors.
Meter Bridge or Slide Wire Bridge:
The meter bridge, also known as the slide wire bridge, is a practical form of the Wheatstone bridge. It consists of a long uniform wire of uniform cross-section called the meter wire, which is stretched over a wooden board. The meter wire is usually made of manganin or constantan, which have low temperature coefficients of resistance.
Construction:
The meter bridge consists of a uniform wire AB, which is connected to a galvanometer G at its midpoint. The wire is divided into two parts by a gap at the center, where a resistance box is connected. A jockey J is used to make contact with the wire and slide along its length.
Working Principle:
To measure an unknown resistance using the meter bridge, the jockey is initially placed at the midpoint of the wire (point O). The resistance box is adjusted until the galvanometer shows zero deflection. At this point, the bridge is in a balanced condition.
The principle behind the working of the meter bridge is that when the bridge is balanced, the ratio of the lengths of the two arms of the wire (AO and OB) is equal to the ratio of the resistances (R1 and R2) in the two arms. Mathematically, this can be represented as:
R1/R2 = AO/OB
By measuring the lengths AO and OB, the ratio R1/R2 can be determined. Since one of the resistances (R1 or R2) is known, the unknown resistance can be calculated.
Advantages of Meter Bridge:
1. High Accuracy: The meter bridge offers high accuracy in measuring resistance as it is based on the principle of a balanced bridge.
2. Simple Construction: The meter bridge is relatively simple in construction and easy to use.
3. Low Cost: The materials required for constructing a meter bridge are easily available and inexpensive.
4. Versatility: The meter bridge can be used to measure a wide range of resistances, making it a versatile instrument.
In conclusion, the meter bridge or slide wire bridge is a practical form of the Wheatstone bridge and is used to measure unknown resistances accurately.
- a)0.04 V
- b)40.0 V
- c)0.4 V
- d)4.0 V
Correct answer is 'D'. Can you explain this answer?
|
Keerthana Iyer answered |
Given:
- e.m.f of first cell = 2V
- Balance point with first cell = 30cm
- Balance point with second cell = 60cm
To find:
- e.m.f of the second cell
Explanation:
To understand this problem, we need to understand the working principle of a potentiometer. A potentiometer is a device used to measure potential difference (or voltage) accurately. It consists of a long wire of uniform cross-section, a jockey, and a galvanometer.
When the jockey is moved along the wire, the galvanometer shows a deflection. At a certain point, the deflection becomes zero, indicating that the potential difference across that point is equal to the potential difference across the terminals of the cell being tested.
In this problem, the balance point with the first cell is at 30cm. This means that the potential difference across 30cm of the wire is equal to the e.m.f of the first cell, which is 2V.
Now, when the cell is replaced by another cell, the balance point shifts to 60cm. This means that the potential difference across 60cm of the wire is equal to the e.m.f of the second cell.
To find the e.m.f of the second cell, we can use the concept of proportionality. The potential difference across the wire is directly proportional to the length of the wire. So, we can set up the following proportion:
Potential difference across 30cm / Length of the wire = Potential difference across 60cm / Length of the wire
Since the length of the wire is the same on both sides of the equation, we can simplify the proportion to:
Potential difference across 30cm = Potential difference across 60cm
Substituting the given values:
2V = Potential difference across 60cm
Therefore, the e.m.f of the second cell is 2V.
Answer:
The e.m.f of the other cell is 4.0V (Option D).
- e.m.f of first cell = 2V
- Balance point with first cell = 30cm
- Balance point with second cell = 60cm
To find:
- e.m.f of the second cell
Explanation:
To understand this problem, we need to understand the working principle of a potentiometer. A potentiometer is a device used to measure potential difference (or voltage) accurately. It consists of a long wire of uniform cross-section, a jockey, and a galvanometer.
When the jockey is moved along the wire, the galvanometer shows a deflection. At a certain point, the deflection becomes zero, indicating that the potential difference across that point is equal to the potential difference across the terminals of the cell being tested.
In this problem, the balance point with the first cell is at 30cm. This means that the potential difference across 30cm of the wire is equal to the e.m.f of the first cell, which is 2V.
Now, when the cell is replaced by another cell, the balance point shifts to 60cm. This means that the potential difference across 60cm of the wire is equal to the e.m.f of the second cell.
To find the e.m.f of the second cell, we can use the concept of proportionality. The potential difference across the wire is directly proportional to the length of the wire. So, we can set up the following proportion:
Potential difference across 30cm / Length of the wire = Potential difference across 60cm / Length of the wire
Since the length of the wire is the same on both sides of the equation, we can simplify the proportion to:
Potential difference across 30cm = Potential difference across 60cm
Substituting the given values:
2V = Potential difference across 60cm
Therefore, the e.m.f of the second cell is 2V.
Answer:
The e.m.f of the other cell is 4.0V (Option D).
An electric bulb is rated 220 V and 100 W. Power consumed by it operated on 110 V is- a)90 W
- b)75 W
- c)25 W
- d)50 W
Correct answer is option 'C'. Can you explain this answer?
An electric bulb is rated 220 V and 100 W. Power consumed by it operated on 110 V is
a)
90 W
b)
75 W
c)
25 W
d)
50 W
|
Preeti Iyer answered |
P = V^2/R
If V = 220 V we have
100 W = 220^2/R
R = 220^2/100 Ω = 484 Ω. This is the resistance of the bulb.
When V = 110 V, power consumed = V^2/R = 110^2 /484 = 25 W.
The Wheatstone bridge and its balance condition provides a practical method for the determination of- a)Unknown voltage
- b)Unknown current
- c)Unknown Resistance
- d)Unknown Resistivity
Correct answer is option 'C'. Can you explain this answer?
The Wheatstone bridge and its balance condition provides a practical method for the determination of
a)
Unknown voltage
b)
Unknown current
c)
Unknown Resistance
d)
Unknown Resistivity
|
|
Preeti Iyer answered |
A wheatstone bridge is an electrical bridge consisting of two branches of a parallel circuit joined by a galvanometer and used for determining the value of an unknown resistance in one of the branches.
Sensitivity of potentiometer can be increased by- a)Increasing voltage
- b)Decreasing length of potentiometer wire
- c)Increasing the length of potentiometer wire and reducing current.
- d)Increasing current in circuit
Correct answer is option 'C'. Can you explain this answer?
Sensitivity of potentiometer can be increased by
a)
Increasing voltage
b)
Decreasing length of potentiometer wire
c)
Increasing the length of potentiometer wire and reducing current.
d)
Increasing current in circuit
|
|
Preeti Iyer answered |
Sensitivity of potentiometer can be increased by increasing the length of the potentiometer wire and by reducing the current in the circuit by using a rheostat. Both the methods help in decreasing the potential gradient, and thereby increasing the resistivity.
Two bulbs A and B are rated 100 W, 120 V and 10 W, 120 V respectively. They are connected across a 120 V source in series. Calculate the current through each bulb.- a)0.08 A
- b)0.0083 A
- c)0.083 A
- d)8 A
Correct answer is option 'C'. Can you explain this answer?
Two bulbs A and B are rated 100 W, 120 V and 10 W, 120 V respectively. They are connected across a 120 V source in series. Calculate the current through each bulb.
a)
0.08 A
b)
0.0083 A
c)
0.083 A
d)
8 A
|
|
Prateek Jain answered |
Bulb A,
PA=100w
VA=120v
We know that, P=VI
So, I=P/V=100w/120v=5/6A=0.83A
Bulb B,
PB=100w
VB=120v
We know that, P=VI
So, I=P/V=10w/120v=1/12A=0.083A
Bulb A will consume more energy than bulb B when they are connected in parallel.
PA=100w
VA=120v
We know that, P=VI
So, I=P/V=100w/120v=5/6A=0.83A
Bulb B,
PB=100w
VB=120v
We know that, P=VI
So, I=P/V=10w/120v=1/12A=0.083A
Bulb A will consume more energy than bulb B when they are connected in parallel.
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