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Question: An inductor of inductance L and resistor of resistance R are joined in series and connected by a sou...

An inductor of inductance L and resistor of resistance R are joined in series and connected by a source of frequency ω∘\omega_\circ. Power dissipated in the circuit is:

Explanation

Solution

Power is defined as the rate at which work is done. It is a scalar quantity as it is the ratio of work done or heat dissipated to the total time taken i.e. P=WorktimeP = \dfrac{Work}{time}. Work is defined as the quantity of energy spent on a system to make it. It is also the measure of the body displaced by a certain force. It is also a scalar quantity.
Formula used:
P=VIcosϕP = VI cos\phi

Complete answer:
Since electrical instruments are connected in series, the current in both the instruments at a particular time will be the same.
Also, we know;
V = IZ
Or I=VZI = \dfrac VZ
Also, using impedance triangle, we can write cosϕcos\phi as:

cosϕ=RZcos\phi = \dfrac{R}{Z}
Also, Z=R2+(ω∘L)2Z = \sqrt{R^2 + (\omega_\circ L)^2}
Now, putting the values of ‘I’ and cosϕcos\phi in P=VIcosϕP = VI cos\phi, we get;
P=VVZRZ=V2RZ2P = V\dfrac{V}{Z}\dfrac RZ = \dfrac{V^2R}{Z^2}
As Z2=R2+(ω∘L)2Z^2 = R^2 +(\omega_\circ L)^2
Thus, P=V2RR2+(ω∘L)2P = \dfrac{V^2R}{R^2+{(\omega_\circ L)}^2}
Hence the power dissipated in the circuit would be: P=V2RR2+(ω∘L)2P = \dfrac{V^2R}{R^2+{(\omega_\circ L)}^2}.

Additional Information:
The fundamental expression of powerP=WorktimeP = \dfrac{Work}{time}.
The S.I. units of power=Jsec or J/s= \dfrac{J}{sec}\ or\ J/s.
This is also known as Watt.
1 Watt is defined when 1 Joule of work is done for 1 sec. The commercial unit of energy is kilowatt-hour. One might get confused that as watt is mentioned in the unit, so it must be the unit of power. But this is not true. Observe the presence of the hour either. This suggests that it is the product of power and time, which is work.

Note:
We can rewrite the power expression in terms of current also. We have to act according to the options given in the question. If no options are given, any expression is correct. Students must not get confused about the formula of power. P=VIP = VI is the expression in case of power dissipation in DC circuits whereas P=VIcosϕP = VI cos\phi is the power dissipation in case of AC circuits.