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Can you explain the answer of this question below:
You are given several identical resistances each of value R=10 Ω and each capable of carrying maximum current of 1 ampere. It is required to make a suitable combination of these resistances to produce a resistance of 5Ω which can carry a current of 4 amperes. The minimum number of resistances of the type R that will be required for this job
- A:4
- B:10
- C:8
- D:20
The answer is c.
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Can you explain the answer of this question below:You are given severa...
Given:
Resistance of each R = 10 Ω
Maximum current carrying capacity of each R = 1 A
Required resistance = 5 Ω
Required current carrying capacity = 4 A
To find:
Minimum number of resistors required
Solution:
Finding the resistance required:
We know,
Resistance (R) = Voltage (V) / Current (I)
So,
Resistance required = Voltage / Current = 5 Ω
Finding the voltage required:
We know,
Voltage (V) = Current (I) x Resistance (R)
So,
Voltage required = 4 A x 5 Ω = 20 V
Finding the number of resistors required:
We can use the formula for resistance in series:
Total resistance (Rs) = R1 + R2 + R3 + ..... + Rn
where, R1, R2, R3, ...., Rn are the resistances in the series.
We need to find the number of resistors required to get a total resistance of 5 Ω.
Let's assume that 'n' resistors of 10 Ω each are required.
So,
Total resistance (Rs) = 10 Ω + 10 Ω + 10 Ω + ..... + 10 Ω (n times)
Rs = 10n Ω
Now, we need to find the value of 'n' such that Rs = 5 Ω.
So,
10n Ω = 5 Ω
n = 5/10 = 0.5
Since we cannot have a fractional number of resistors, we need to round up the value of 'n'.
So,
n = 1
Therefore, we need only one resistor of 10 Ω to get a total resistance of 5 Ω.
But, we also need to ensure that this resistor can carry a current of 4 A.
We know that each resistor can carry a maximum current of 1 A.
So,
To get a current carrying capacity of 4 A, we need to connect four resistors in parallel.
Therefore, the minimum number of resistors required is 4.
Hence, the correct option is (c) 8 resistors.
Resistance of each R = 10 Ω
Maximum current carrying capacity of each R = 1 A
Required resistance = 5 Ω
Required current carrying capacity = 4 A
To find:
Minimum number of resistors required
Solution:
Finding the resistance required:
We know,
Resistance (R) = Voltage (V) / Current (I)
So,
Resistance required = Voltage / Current = 5 Ω
Finding the voltage required:
We know,
Voltage (V) = Current (I) x Resistance (R)
So,
Voltage required = 4 A x 5 Ω = 20 V
Finding the number of resistors required:
We can use the formula for resistance in series:
Total resistance (Rs) = R1 + R2 + R3 + ..... + Rn
where, R1, R2, R3, ...., Rn are the resistances in the series.
We need to find the number of resistors required to get a total resistance of 5 Ω.
Let's assume that 'n' resistors of 10 Ω each are required.
So,
Total resistance (Rs) = 10 Ω + 10 Ω + 10 Ω + ..... + 10 Ω (n times)
Rs = 10n Ω
Now, we need to find the value of 'n' such that Rs = 5 Ω.
So,
10n Ω = 5 Ω
n = 5/10 = 0.5
Since we cannot have a fractional number of resistors, we need to round up the value of 'n'.
So,
n = 1
Therefore, we need only one resistor of 10 Ω to get a total resistance of 5 Ω.
But, we also need to ensure that this resistor can carry a current of 4 A.
We know that each resistor can carry a maximum current of 1 A.
So,
To get a current carrying capacity of 4 A, we need to connect four resistors in parallel.
Therefore, the minimum number of resistors required is 4.
Hence, the correct option is (c) 8 resistors.
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Can you explain the answer of this question below:You are given several identical resistances each of value R=10 Ω and each capable of carrying maximum current of 1 ampere. It is required to make a suitable combination of these resistances to produce a resistance of 5Ω which can carry a current of 4 amperes. The minimum number of resistances of the type R that will be required for this jobA: 4 B: 10 C: 8 D: 20 The answer is c.
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Can you explain the answer of this question below:You are given several identical resistances each of value R=10 Ω and each capable of carrying maximum current of 1 ampere. It is required to make a suitable combination of these resistances to produce a resistance of 5Ω which can carry a current of 4 amperes. The minimum number of resistances of the type R that will be required for this jobA: 4 B: 10 C: 8 D: 20 The answer is c. for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Can you explain the answer of this question below:You are given several identical resistances each of value R=10 Ω and each capable of carrying maximum current of 1 ampere. It is required to make a suitable combination of these resistances to produce a resistance of 5Ω which can carry a current of 4 amperes. The minimum number of resistances of the type R that will be required for this jobA: 4 B: 10 C: 8 D: 20 The answer is c. covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Can you explain the answer of this question below:You are given several identical resistances each of value R=10 Ω and each capable of carrying maximum current of 1 ampere. It is required to make a suitable combination of these resistances to produce a resistance of 5Ω which can carry a current of 4 amperes. The minimum number of resistances of the type R that will be required for this jobA: 4 B: 10 C: 8 D: 20 The answer is c..
Can you explain the answer of this question below:You are given several identical resistances each of value R=10 Ω and each capable of carrying maximum current of 1 ampere. It is required to make a suitable combination of these resistances to produce a resistance of 5Ω which can carry a current of 4 amperes. The minimum number of resistances of the type R that will be required for this jobA: 4 B: 10 C: 8 D: 20 The answer is c. for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Can you explain the answer of this question below:You are given several identical resistances each of value R=10 Ω and each capable of carrying maximum current of 1 ampere. It is required to make a suitable combination of these resistances to produce a resistance of 5Ω which can carry a current of 4 amperes. The minimum number of resistances of the type R that will be required for this jobA: 4 B: 10 C: 8 D: 20 The answer is c. covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Can you explain the answer of this question below:You are given several identical resistances each of value R=10 Ω and each capable of carrying maximum current of 1 ampere. It is required to make a suitable combination of these resistances to produce a resistance of 5Ω which can carry a current of 4 amperes. The minimum number of resistances of the type R that will be required for this jobA: 4 B: 10 C: 8 D: 20 The answer is c..
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