A 10 ohm resistor is connected in parallel across series combination of 4 ohms and 6 ohms resistors. calculate the amount of heat produced per second, in 4 ohms resistor if 10 joules heat is produced per second in 10 ohms resistor

OK. We can do this !

-- We need to know the current through the 4-ohm resistor.

-- But the only thing we know is the power dissipated by the 10-ohm
resistor. Can we use that to find the current through the 4-ohm resistor ?

You bet we can !

-- The effective resistance of the series combination is (4+6) = 10 ohms.

-- Both parallel branches have the same resistance, so the current
is the same in both branches.

-- The power dissipated by a resistor is P = I² · R
For the 10-ohm resistor,

   (10 joules/sec) = (I²) · 10-ohms

   I = √(10/10) = 1 ampere

-- The current in the series combination is the same as the current
in the 10-ohm resistor . . . 1 ampere.

-- So the power dissipated by the 4-ohm resistor in the series branch is

   P = I² · R = (1 amp)² · (4 ohms) = 4 watts.

adityaso adityaso · Answer 2 · 2020-01-25T10:14:57+01:00

The amount of heat produced per second in the [tex]4\,\Omega[/tex] resistor is [tex]\boxed{4\text{ J/s}}[/tex].

Explanation:

Given:

The resistance in the given circuit are [tex]4\,\Omega, 6\,\Omega\text{ and }10\,\Omega[/tex].

The power of the [tex]10\,\Omega[/tex] resistance is [tex]10\text{ J/s}[/tex].

Concept:

The amount of heat produced in a resistor per second is defined as the power produced in the resistor.

Write the expression for the equivalent resistance of the series combination of two resistors.

[tex]R_{s}=R_{1}+R_{2}[/tex]

Substitute the value of the resistance in the series combination.

[tex]\begin{aligned}R_{s}&=4\,\Omega+6\,\Omega\\&=10\,\Omega\end{aligned}[/tex]

The another [tex]10\,\Omega[/tex] resistance in parallel with the resistance of the series combination. Therefore, the current through both the arms containing the resistance will be same.

Write the expression for the power in the resistance.

[tex]\boxed{P=I^2R}[/tex]

Substitute [tex]10\,\Omega[/tex] for [tex]R[/tex] and [tex]10\text{ J/s}[/tex] for [tex]P[/tex] in above expression.

[tex]\begin{aligned}10&=I^2\times10\\I&=\sqrt{1}\\&=1\text{ A}\end{aligned}[/tex]

The current through the [tex]4\,\Omega[/tex] resistance will also be [tex]1\text{ A}[/tex]. So, the power or the heat produced per second in the [tex]4\,\Omega[/tex] resistance will be.

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

Substitute [tex]4\,\Omega[/tex] for [tex]R'[/tex] and [tex]1\text{ A}[/tex] for [tex]I[/tex] in above expression.

[tex]\begin{aligned}P'&=(1)^2(4\,\Omega)\\&=4\text{ J/s}\end{aligned}[/tex]

Thus, The amount of heat produced per second in the [tex]4\,\Omega[/tex] resistor is [tex]\boxed{4\text{ J/s}}[/tex].

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Answer Details:

Grade: High School

Subject: Physics

Chapter: Current Electricity

Keywords:

10 ohm, resistor, parallel connection, series combination, connected, heat, per second, power, current, 10 J/s, 4 ohms, 6 ohms.