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.
Can you explain the answer of this question below:You are given severa...
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