Power dissipated across the 8 Ω resistor in the circuit shown here is 2 watt. The power dissipated in watt units across the 3 Ω resistor is :-

Question

Power dissipated across the 8 Ω resistor in the circuit shown here is 2 watt. The power dissipated in watt units across the 3 Ω resistor is :- (A) 3.0 (B) 2.0 (C) 1.0 (D) 0.5

Tardigrade Admin · Accepted Answer

As voltage drop across 8 Ω=√2 × 8=4 V (∵ P =( V 2/ R )) Therefore voltage drop across 3 Ω=3 V [∵ 4 V is divided in ratio of resistances between 1 Ω and 3 Ω] Hence power dissipated in 3 Ω=((3)2/3)=3 watt

Q. Power dissipated across the $8 Ω$ resistor in the circuit shown here is $2$ watt. The power dissipated in watt units across the $3 Ω$ resistor is :-

3.0

2.0

1.0

0.5

As voltage drop across
$8Ω = 2 \times 8 = 4 V (∵ P = \frac{V ^{2}}{R})$
Therefore voltage drop across $3Ω = 3 V$
$[∵ 4 V$ is divided in ratio of resistances between $1Ω$ and $3Ω]$
Hence power dissipated in $3Ω = \frac{( 3 ) ^{2}}{3} = 3$ watt

Q. Power dissipated across the 8Ω resistor in the circuit shown here is 2 watt. The power dissipated in watt units across the 3Ω resistor is :-

3.0

2.0

1.0

0.5

As voltage drop across 8Ω=2×8​=4V(∵P=RV2​) Therefore voltage drop across 3Ω=3V [∵4V is divided in ratio of resistances between 1Ω and 3Ω] Hence power dissipated in 3Ω=3(3)2​=3 watt

Q. Power dissipated across the $8 Ω$ resistor in the circuit shown here is $2$ watt. The power dissipated in watt units across the $3 Ω$ resistor is :-

As voltage drop across
$8Ω = 2 \times 8 = 4 V (∵ P = \frac{V ^{2}}{R})$
Therefore voltage drop across $3Ω = 3 V$
$[∵ 4 V$ is divided in ratio of resistances between $1Ω$ and $3Ω]$
Hence power dissipated in $3Ω = \frac{( 3 ) ^{2}}{3} = 3$ watt