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A 100 V Battery is Connected to the Electric Network as Shown. If the Power Consumed in the 2 ω Resistor is 200 W, Determine the Power Dissipated in the 5 ω Resistor. - Physics

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ConceptElectric Current

Question

 A 100 V battery is connected to the electric network as shown. If the power consumed in the 2 Ω resistor is 200 W, determine the power dissipated in the 5 Ω resistor.

Solution

 Total current\[I = \sqrt{\frac{P}{R}} = \sqrt{\frac{200}{2}} = 10A\]

Equivalent circuit:

Potential difference across AB = potential difference across CD

\[\Rightarrow I_1 R_1 = I_2 R_2 \]

\[ \Rightarrow 10 I_1 - 40 I_2 = 0\]

\[ \Rightarrow I_1 - 4 I_2 = 0 . . . (1)\]

\[ \Rightarrow I_1 + I_2 = I\]

\[ \Rightarrow I_1 + I_2 = 10 . . . (2)\]

\[ \Rightarrow I_1 = 8A\]

Power dissipated in the 5 Ω resistor = 5 \[\times\]  (current through 5 Ω resistor)2
∴ P = 320 W

  Is there an error in this question or solution?

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Solution A 100 V Battery is Connected to the Electric Network as Shown. If the Power Consumed in the 2 ω Resistor is 200 W, Determine the Power Dissipated in the 5 ω Resistor. Concept: Electric Current.
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