In the circuit, the equivalent resistance between A and B is (Each resistance given in the diagram is R) :

F1 Prabhu 5.1.20 Pallavi D3.1

  1. R Ω
  2. \(\dfrac{R}{2}\Omega\)
  3. \(\dfrac{R}{4}\Omega\)
  4. 2 R Ω

Answer (Detailed Solution Below)

Option 1 : R Ω
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CUET General Awareness (Ancient Indian History - I)
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Detailed Solution

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CONCEPT:

  • Equivalent resistance: It shows where the aggregate resistance connected either in parallel or series is calculated.
    • Essentially, the circuit is designed either in Series or Parallel.
    • Electrical resistance shows how much energy one needs when you move the charges/current through your devices.

Formula for resistors connected in series / parallel:

  1. R = R1 + R2 + R3........(Series)
  2. 1/R = 1/R1 + 1/R2 + 1/R3 ......(Parallel)

EXPLANATION:

Given that,

F1 Prabhu Ravi 12.05.21 D1

Value of all resistance is R and hence R1 = R2 = R3 = R4 = R5 = R

Identify the Junction points Between A and B and redraw the circuit.

F1 Prabhu Ravi 12.05.21 D2

Clearly, we can see R2 is forming a Wheatstone bridge.

R1 / R3 = R / R = 1

R5 / R4 = R /R = 1

So, R1 /R3 = R5 / R4 So, it is a balanced Wheatstone bridge and hence no current will be in the PS branch.

R1 and R5 are in series equivalent = R + R = 2R

and R3 and R4 are in series, equivalent = R + R = 2R

Both are in parallel

So,

Req =(2R)2 / 2R +2R = R

So, the equivalent resistance is R

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