Understanding the RMS (Root Mean Square) Value:
The root mean square (rms) value \( I_{\text{rms}} \) of a current \( i = I_0 + I_1 \sin(\omega t + \phi) \) is given by:
\[ I_{\text{rms}} = \sqrt{(I_0)^2 + \frac{(I_1)^2}{2}} \] where \( I_0 \) is the DC component and \( I_1 \) is the amplitude of the AC component.
Identify \( I_0 \) and \( I_1 \):
In this case:
\[ I_0 = 6 \, \text{A} \quad \text{and} \quad I_1 = \sqrt{56} \, \text{A} \]
Calculate the RMS Value:
Substitute \( I_0 = 6 \) and \( I_1 = \sqrt{56} \) into the rms formula:
\[ I_{\text{rms}} = \sqrt{(6)^2 + \frac{(\sqrt{56})^2}{2}} \] \[ = \sqrt{36 + \frac{56}{2}} \] \[ = \sqrt{36 + 28} \] \[ = \sqrt{64} = 8 \, \text{A} \]
Conclusion:
The rms value of the current is \( 8 \, \text{A} \).



In the given figure, the blocks $A$, $B$ and $C$ weigh $4\,\text{kg}$, $6\,\text{kg}$ and $8\,\text{kg}$ respectively. The coefficient of sliding friction between any two surfaces is $0.5$. The force $\vec{F}$ required to slide the block $C$ with constant speed is ___ N.
(Given: $g = 10\,\text{m s}^{-2}$) 
Two circular discs of radius \(10\) cm each are joined at their centres by a rod, as shown in the figure. The length of the rod is \(30\) cm and its mass is \(600\) g. The mass of each disc is also \(600\) g. If the applied torque between the two discs is \(43\times10^{-7}\) dyne·cm, then the angular acceleration of the system about the given axis \(AB\) is ________ rad s\(^{-2}\).
