Question

A composite resistance of 50 ohm which can carry a current of 4A is to be made from resistances each of resistance 100ohm which can carry a current of 1A. the minimum number of resistances to be used is: a) 4 b) 8 c) 12 d) 16.

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Answer 1

Answer:

• Option (B) 8

Explanation:

Given that,

●Composite resistance = 50 ohm

●Current = 4 A

●Each resistance = 100 ohm

●Each resistance carry a current = 1 A

Suppose,

In one pair, two resistance connected in series.

So, the value of one pair of resistance is 200 ohm because each resistance is 100 ohm.

Now, four pairs of resistance connected in parallel.

We need to calculate the equivalent resistance ;

Using the formula of parallel

↪1/Req = 1/200+1/200+1/200+1/200

↪1/Req = 1/50

↪Req = 50Ω $\red\bigstar$

Hence, The minimum number of resistance to be used is 8.

For information:

• I = Q/t

• V = W/Q

• R = ρl/A

• P = VI

• σ = 1/ρ

Answer 2

Given :-

• Equivalent resistance of the combination = 50 Ω
• total current flowing = 4A
• Resistance of each resistor  (R) = 100 Ω

How to solve :-

• Equivalent resistance will come out to be 50Ω only when four rows each having having a resistance of 200Ω are connected in parallel combination
• This means that each row will have two resistors each having resistance of 100 Ω

Calculation :-

Equivalent resistance of two resistors connected in series is given by  ,

➛ Rs = R1 + R2

➛ Rs = 2R        

➛ Rs = 200 Ω

Equivalent resistance of resistors connected in parallel combination is given by ,

➛ 1 / Rp = 1 / Rs + 1 / Rs + 1 / Rs + 1 / Rs

➛ 1 / Rp = 4 / Rs

➛ Rp = Rs / 4

➛ Rp = 200 / 4

➛ Rp = 50 Ω

The minimum number of resistances used is 8