In the circuit shown below the power loss in 5- resistor is W-

Applying KCL at point 'A' 10+V_1 101 2+10+V_1 4V_15=0 10 10+V_1 2+2+ 3V_15=0 V_1=0 Current, i= 10 3V_15=2A Power loss in 5Ω resistor i^2R=(2)^2×5= 20 W nb ...

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Circuit Theory

Super position theorem(SPT)

In the circui...

Question

In the circuit shown below the power loss in $5 Ω$ resistor is _________W.

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Solution

The correct option is A 20

Applying KCL at point 'A'
$\frac{10 + V_{1} - 10}{1} - 2 + \frac{10 + V_{1} - 4 V_{1}}{5} = 0$
$10 - 10 + V_{1} - 2 + 2 + \frac{- 3 V_{1}}{5} = 0$
$V_{1} = 0$
Current, $i = \frac{10 - 3 V_{1}}{5} = 2 A$
Power loss in $5 Ω$ resistor $i^{2} R = (2)^{2} \times 5 = 20 W$

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In the circuit shown below the power loss in 5Ω resistor is _________W.

The correct option is A 20 Applying KCL at point 'A' 10+V1−101−2+10+V1−4V15=0 10−10+V1−2+2+−3V15=0 V1=0 Current, i=10−3V15=2A Power loss in 5Ω resistor i2R=(2)2×5=20W

In the circuit shown below the power loss in $5 Ω$ resistor is _________W.

The correct option is A 20

Applying KCL at point 'A'
$\frac{10 + V_{1} - 10}{1} - 2 + \frac{10 + V_{1} - 4 V_{1}}{5} = 0$
$10 - 10 + V_{1} - 2 + 2 + \frac{- 3 V_{1}}{5} = 0$
$V_{1} = 0$
Current, $i = \frac{10 - 3 V_{1}}{5} = 2 A$
Power loss in $5 Ω$ resistor $i^{2} R = (2)^{2} \times 5 = 20 W$