# Alternating Current - Class 12 - Physics

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## Extra Questions - Alternating Current | NCERT | Physics | Class 12

When a system is taken from state '$a$' to state '$b$' along the path '$ac b$', it is found that a quantity of heat $Q = 200 , \mathrm{J}$ is absorbed by the system and work $W = 80 , \mathrm{J}$ is done by it. Along the path '$ad b$', $Q = 144 , \mathrm{J}$. The work done along the path '$ad b$' is

A) $6 , \mathrm{J}$

B) $12 , \mathrm{J}$

C) $18 , \mathrm{J}$

D) $24 , \mathrm{J}$

The correct option is **D) $24 , \mathrm{J}$**.

To solve for the work done along the path '$adb$', we utilize the **First Law of Thermodynamics** which states that the change in internal energy $\Delta U$ of a system is equal to the heat added to the system minus the work done by the system. It's important to remember that **internal energy ($\Delta U$) is a state function**, which means it is path independent and solely dependent on the initial and final states of the system.

From the given data for the path '$acb$': $$ Q_{\text{acb}} = 200 , \mathrm{J} $$ $$ W_{\text{acb}} = 80 , \mathrm{J} \quad (\text{work done by the system}) $$ The change in internal energy from state $a$ to state $b$ ($\Delta U_{ab}$) can be calculated as: $$ \Delta U_{ab} = Q_{\text{acb}} - W_{\text{acb}} = 200 , \mathrm{J} - 80 , \mathrm{J} = 120 , \mathrm{J} $$

For the path '$adb$', where $Q_{adb} = 144 , \mathrm{J}$, we apply the First Law of Thermodynamics again: $$ \Delta U_{ab} = Q_{adb} - W_{adb} $$ Therefore, solving for $W_{adb}$: $$ W_{adb} = Q_{adb} - \Delta U_{ab} = 144 , \mathrm{J} - 120 , \mathrm{J} = 24 , \mathrm{J} $$

Thus, the work done along the path '$adb$' is **24 Joules**.

A box containing a resistance $R$, inductance $L$, and a capacitor of capacitance $C$ is connected in series to an alternating source of angular frequency $4 \mathrm{rad/s}$ and a capacitor of capacitance $C$ as shown in the circuit. The box has a power factor of $\frac{1}{\sqrt{3}}$, and the circuit has an overall power factor of 1. The impedance of the box is:

A) $\frac{\sqrt{3}}{4 \sqrt{2} C}$

B) $\frac{1}{4 C}$

C) $\frac{1}{4 \sqrt{2} C}$

D) $\frac{5}{4 \sqrt{2} C}$

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The dimensions of $1 / \mu_{0} \varepsilon_{0}$, where symbols have their usual meanings, are

A $\left[L^{-1} \mathrm{~T}\right]$

B $\left[L^{-2} T^{2}\right]$

C $\left[\mathrm{L}^{2} \mathrm{~T}^{-2}\right]$

D $\left[L T^{-1}\right]$