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How much power is the 20Ω resistor dissipating?
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The 10Ω resistor in the figure (Figure 1) is dissipating 35 W of power.
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Refer to picture

How much power is the 20Ω resistor dissipating The 10Ω resistor in the figure Figure 1 is dissipating 35 W of power Refer to picture class=

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skyluke89
The power dissipated by a resistor is given by:
[tex]P=I^2 R[/tex] (1)
where I is the current through the resistor and R is the resistance.

We are told that the resistor of [tex]10 \Omega[/tex] dissipates 35 W of power. By rearranging the previous equation (1), and by using these data, we can find the current flowing through the resistor of [tex]10 \Omega[/tex]:
[tex]I_L = \sqrt{\frac{P}{R} }= \sqrt{ \frac{35 W}{10 \Omega} } =1.87 A [/tex]

This is the current flowing in the left branch of the circuit: we need to find the current flowing in the right branch, [tex]I_R[/tex].
The equivalent resistance on the left branch is the sum of the two resistances, since they are in series:
[tex]R_{L}=10 \Omega + 5 \Omega = 15 \Omega[/tex]
We know that this resistance is in parallel with the resistor on the right, [tex]R_R = 20 \Omega[/tex], so they have same potential difference:
[tex]V_L=V_R[/tex]
which can be rewritten, by using Ohm's law:
[tex]I_L R_L = I_R R_R[/tex]
Rearranging the equation, we find the current through the right branch:
[tex]I_R= \frac{I_L R_L}{R_R}= \frac{(1.87 A)(15 \Omega)}{20 \Omega}=1.40 A [/tex]

And now we can find the power dissipated through the resistor on the right:
[tex]P=I^2R=(1.40 A)^2(20 \Omega)=39.2 W[/tex]

The power dissipated through the resistor on the right is [tex]39.2\;\text{W}[/tex]

Explanation:

Given information:

The resistance 20Ω , 10Ω and 5Ω

The power dissipated by a resistor is given by

[tex]P=I^2R[/tex]

According to the question the resistor of 10Ω dissipates 35 W of power.

So,

[tex]I_L=\sqrt{ \frac{P}{R} }\\I_L=\sqrt{35/10} \\I_L=1.87\;A[/tex]

This is the current flowing in the left branch of the circuit.

Similarly for the right branch:

[tex]R_L=15[/tex] Ω

[tex]R_R=20[/tex] Ω

Now, according to the ohm's law:

[tex]I_LR_L=I_RR_R[/tex]

on putting the values in above equation:

[tex]I_R=(1.87\times 15)/20\\I_R=1.40\;A[/tex]

Hence, the power dissipated through the resistor on the right

[tex]P=I^2R\\P=1.40^2\times20\\P=39.2\;W[/tex]

The power dissipated through the resistor on the right is [tex]39.2\;\text{W}[/tex]

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