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A current of 3 amp. flows through the 2Ω resistor shown in the circuit. The power dissipated in the 5Ω resistor is -

A current of 3 amp. flows through the 2Ω resistor shown in the circuit. The power dissipated in the 5Ω resistor is - Correct Answer 5 watt

CONCEPT:

  • According to Ohms Law, Potential Difference (V) across the resistor is the product of current  (I) and Resistance (R).
  • The parallel combination of resistors: When Resistors are connected across the same potential difference they are said to be Parallel.
  • Series combination of the resistors: When Resistors are connected across the same path of the current, so that the same current is passing through them, they are said to be in series. 
    • Equivalent Resistance of two or more resistors in series is the Algebraic sum of the Resistances.

Electric Power (P) across Resistor have Current (I) across it and resistance (R) is given as:

​P = I2 R

​CALCULATION:

For the sake of understanding, the resistors have been named accordingly.

[ alt="F2 Jitendra Madhu 27.10.20 D5" src="//storage.googleapis.com/tb-img/production/20/10/F2_Jitendra_Madhu_27.10.20_D5.png" style="width: 222px; height: 163px;">

From the above figure, R3 and R5 are in series. This series combination is in Parallel with both R1 and R2.

So, the Potential difference across R1 is the same as the Potential difference across R3 + R4

Given, Current through R1 (2Ω) = 3 A

Potential Difference across R1 = 3A × 2Ω = 6V --- (1)

Equaivalent Resistance across R3 + R4 = 1Ω + 5Ω = 6Ω

Potential Difference Across (R3 + R4) = Potential Difference across R1 = 6V 

Current through (R3 + R4) = 6V / 6Ω = 1A

Since R3 and R4 are in series, the current through them is the same that is 1A

Now, we have to find Power dissipated by 5 Ω resistor.

Power = I2 R5 = (1A)2 × 5Ω  = 5 watt.

So, 5 watt is the correct answer.

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