Physics, 28.09.2019 04:30 jbearallen9557

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 (figu

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Answer from: LilFreaky666

The power dissipated by a resistor is given by:
P=I^2 R

(1)
where I is the current through the resistor and R is the resistance.

We are told that the resistor of $10 \Omega$ 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 $10 \Omega$ :
$I_L = \sqrt{\frac{P}{R} }= \sqrt{ \frac{35 W}{10 \Omega} } =1.87 A$

This is the current flowing in the left branch of the circuit: we need to find the current flowing in the right branch, I_R

.
The equivalent resistance on the left branch is the sum of the two resistances, since they are in series:
$R_{L}=10 \Omega + 5 \Omega = 15 \Omega$
We know that this resistance is in parallel with the resistor on the right, $R_R = 20 \Omega$ , so they have same potential difference:
V_L=V_R

which can be rewritten, by using Ohm's law:
I_L R_L = I_R R_R

Rearranging the equation, we find the current through the right branch:
$I_R= \frac{I_L R_L}{R_R}= \frac{(1.87 A)(15 \Omega)}{20 \Omega}=1.40 A$

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

Answer from: shira35

The power dissipated through the resistor on the right is $39.2\;\text{W}$

Explanation:

Given information:

The resistance 20Ω , 10Ω and 5Ω

The power dissipated by a resistor is given by

P=I^2R

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

So,

$I_L=\sqrt{ \frac{P}{R} }\\I_L=\sqrt{35/10} \\I_L=1.87\;A$

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

Similarly for the right branch:

R_L=15 Ω

R_R=20 Ω

Now, according to the ohm's law:

I_LR_L=I_RR_R

on putting the values in above equation:

$I_R=(1.87\times 15)/20\\I_R=1.40\;A$

Hence, the power dissipated through the resistor on the right

$P=I^2R\\P=1.40^2\times20\\P=39.2\;W$

The power dissipated through the resistor on the right is $39.2\;\text{W}$

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