AC Power Calculation with Impedance
The initial power P drawn by a resistor R connected to a constant AC voltage source ($V_{rms}$) is:
$P = \frac{V_{rms}^2}{R}$
When an inductance is added in series with the resistor, the total impedance of the circuit becomes Z. The source voltage $V_{rms}$ remains unchanged.
The new RMS current $I'_{rms}$ in this series R-L circuit is calculated using Ohm's law for AC circuits:
$I'_{rms} = \frac{V_{ سےms}}{Z}$
The power dissipated by the resistor in the new circuit is the new power $P'$. Since only the resistor dissipates average power in an AC circuit:
$P' = (I'_{rms})^2 R$
Substitute the expression for $I'_{rms}$:
$P' = \left(\frac{V_{rms}}{Z}\right)^2 R = \frac{V_{rms}^2}{Z^2} R$
From the initial power equation ($P = \frac{V_{rms}^2}{R}$), we can rearrange to find $V_{rms}^2$:
$V_{rms}^2 = P \times R$
Now, substitute this expression for $V_{rms}^2$ into the equation for $P'$:
$P' = \frac{(P \times R)}{Z^2} R$
Simplify the expression:
$P' = P \frac{R^2}{Z^2}$
$P' = P \left(\frac{R}{Z}\right)^2$
Thus, the power drawn by the circuit in the presence of the inductor will be $P \left( \frac{R}{Z} \right)^2$.
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Important Questions from Electricity and Magnetism
Two blocks of masses $m$ and $M$, ($M > m$), are placed on a frictionless table as shown in figure. A massless spring with spring constant $k$ is attached with the lower block. If the system is slightly displaced and released, then
A. The time period of small oscillation of the two blocks is $T = 2\pi \sqrt{\frac{(m+M)}{k}}$
B. The acceleration of the blocks is $a = \frac{kx}{M+m}$ (x = displacement of the blocks from the mean position)
C. The magnitude of the frictional force on the upper block is $\frac{\mu m|x|}{M+m}$
D. The maximum amplitude of the upper block, if it does not slip, is $\frac{\mu (M+m)g}{k}$
E. Maximum frictional force can be $\mu(M + m) g$.
Choose the correct answer from the options given below:
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