Question

Three resistors 8 ohms, 6 ohms and 10 ohms are connected in series to a battery of terminal voltage 24 Volts. Find the power dissipated across the resistor 6 ohm resistor​

Answer 1

★ Answer :-

Power dissipated across 6Ω resistor=6 watt

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★ Explanation :-

It is given that 3 resistors are connected in series connection with a battery of 24 volts. We are asked to find the power dissipated by 6Ω Resistor. So firstly we will find the current in the whole circuit by using total resistance and voltage of battery and we will apply formula for power to solve this problem.

So let's do it!!

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★ Solution :-

Let's find the total resistance!

Total resistance in series connection can be expressed as::

» Rₙ=R₁+R₂+R₃

By substituting values of resistors we get::

» Rₙ=8Ω+6Ω+10Ω

By solving it we get::

» Rₙ=24Ω

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Now, by applying ohms law for whole circuit::

» V=IRₙ

By substituting values of Rₙ and V::

» 24V=I×(24Ω)

By dividing 24 from both sides::

» 24V÷24=24Ω×I÷24

We get::

» 1 A=I

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Now applying formula for power dissipated::

» P=I²R₁

Now substituting values of I and R₁::

» P=(1A)²×(6Ω)

By solving it we get::

» P=1×6 Watt

Final answer::

» P=6 watt

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★ Formulae related to this chapter :-

» P=W/t

» P=E/t

» P=VI

» P=I²R

» P=V²/R

» V=W/Q

» V=IR

» Word done=Heat

» H=VIt

» H=I²Rt

» H=V²t/R

» I=Q/t

» Q=ne

» R=ρL/A

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★  Important Laws :-

» ohm's law::

V=IR

» Joules law of heating::

H=I²Rt

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