You are given several identical resistances each of value R=10 Ω and e...
To carry a current of 4 amperes, we need four paths, each carrying a current of one ampere. Let r be the resistance of each path. These are connected in parallel. Hence, their equivalent resistance will be r/4. According to the given problem
r/4=5
r=20
For this propose two resistances should be connected. There are four such combinations. Hence, the total number of resistance = 4 * 2 =8
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You are given several identical resistances each of value R=10 Ω and e...
Understanding the problem:
To create a resistance of 5Ω that can carry a current of 4 amperes, we need to determine the minimum number of 10Ω resistors that can be combined to achieve this.
Calculating the equivalent resistance:
- To find the total resistance, we can use the formula for resistors in parallel: 1/Req = 1/R1 + 1/R2 + 1/R3 + ...
- Given that each resistor has a value of 10Ω, we can substitute this into the formula: 1/Req = 1/10 + 1/10 + 1/10 + 1/10 = 4/10
- Simplifying, we get: 1/Req = 0.4, which means Req = 1/0.4 = 2.5Ω
Determining the number of resistors needed:
- Since we are using 10Ω resistors, we must calculate how many of these resistors are needed to achieve an equivalent resistance of 2.5Ω.
- To do this, we divide the desired resistance by the resistance of each individual resistor: 2.5/10 = 0.25
- Therefore, we need 0.25 or 1/0.25 = 4 resistors to achieve the desired equivalent resistance.
Conclusion:
Hence, the minimum number of resistors of 10Ω required to create an equivalent resistance of 2.5Ω is 4. Therefore, the correct answer is option (c) 8 resistors.
You are given several identical resistances each of value R=10 Ω and e...
Solution:
To find the minimum number of resistances required, we need to consider the formula for combining resistances in parallel and in series.
Formula for resistances in series:
When resistances are connected in series, the total resistance is the sum of individual resistances.
R_total = R1 + R2 + R3 + ... + Rn
Formula for resistances in parallel:
When resistances are connected in parallel, the reciprocal of the total resistance is the sum of reciprocals of individual resistances.
1/R_total = 1/R1 + 1/R2 + 1/R3 + ... + 1/Rn
Given that we have identical resistances of value R = 10 Ω, we can use these formulas to find the minimum number of resistances required.
Step 1: Calculate the total resistance required using the formula for resistances in parallel.
1/R_total = 1/R1 + 1/R2 + 1/R3 + ... + 1/Rn
1/5 Ω = 1/10 Ω + 1/10 Ω + 1/10 Ω + ... + 1/10 Ω
1/5 Ω = n/10 Ω
n = 2
Step 2: Calculate the total current required using the formula for resistances in series.
I_total = I1 + I2 + I3 + ... + In
4 A = 1 A + 1 A + 1 A + 1 A
4 A = n A
n = 4
Therefore, the minimum number of resistances required is 4 (Option A).
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