3 Marks Question

Question 13 Marks

Obtain the formula of electric current for an AC circuit having only capacitor.

Answer

→As shown in the fig., an AC source is connected to a capacitor.
→Voltage of the AC source,
$v=v_m \sin \omega t$
→A capacitor connected to an AC source, limits or regulates the current, but does not completely prevent the flow of charge.
→The capacitor is alternatively charged and discharged as the current reverses each half cycle.
→Let $q$ be the charge on the capacitor at any time $t$.
→The instantaneous voltage $v$ across the capacitor is,
$\begin{array}{ll}
& v=\frac{q}{ C } \\
\therefore & v_m \sin \omega t=\frac{ q }{ C } \\
\therefore & q=v_m \cdot C \sin \omega t
\end{array}$
→To find the current, we use the relation,
$\begin{aligned}
& i=\frac{d q}{d t} \\
\therefore \quad & \frac{d q}{d t}=v_m C \frac{d}{d t}(\sin \omega t) \\
\therefore \quad & i=v_m \omega C \cos \omega t \\
\therefore \quad i= & v_m \omega C \sin \left(\omega t+\frac{\pi}{2}\right) \\
\therefore \quad i= & \frac{v_m}{\frac{1}{\omega C}} \sin \left(\omega t+\frac{\pi}{2}\right) \\
\therefore \quad & i=i_m \sin \left(\omega t+\frac{\pi}{2}\right)
\end{aligned}$
Where $i_m=\frac{v_m}{\frac{1}{\omega C}}$ (Amplitude of current)
→This equation is similar to the equation
$i_m=\frac{v_m}{R}$ for a purely resistive circuit.
→Thus, the term $\frac{1}{\omega C }$ is similar (or analogus) to resistor in D.C. circuit.
→It is called capacitive reactance and is denoted by $X _{ C }$.
$\therefore \quad X _{ C }=\frac{1}{\omega C } \text { (Unit : ohm }(\Omega) \text { ) }$
→Therefore, the amplitude of electirc current
$i_m=\frac{v_m}{ X _{ C }}$
→From eq. (1) and (2), it can be said that current is $\frac{\pi}{2} rad$ ahead of voltage in phase.

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Question 23 Marks

Obtain the equation of electric current for an AC circuit having only inductor.

Answer

→As shown in fig., an AC source is connected to pure inductor. (A pure inductor means an inductor having negligibly small resistance.)
→Let the voltage across the source be
$v=v_m \sin \omega t$
→Using the kirchhoff's loop rule for the AC circuit shown in the fig.,
$v- L \frac{d i}{d t}=0$

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Question 33 Marks

Accepting the relation $\frac{v_s}{v_p}=\frac{ N _s}{N_s}$ for ideal transformer, and using thé formula of power; derive $\frac{I_p}{I_s}=\frac{v_s}{v_p}=\frac{N_s}{ N _p}$ and from that explain the types of the transformers.

Answer

$\rightarrow$ If the transformer is assumed to be $100 \%$ efficient $($no energy losses$),$ the input power is equal to the output power,
$\therefore I _p U _p= I _s U _s$
$\therefore \frac{ U _s}{ v _p}=\frac{ I _p}{ I _s}$
But $\frac{ U _s}{ v _p}=\frac{ N _s}{N_p}$
$\rightarrow $ But $\frac{v_s}{v_p}=\frac{ N _s}{N_p}$
we get $\frac{v_s}{v_p}=\frac{ N _s}{N_p}=\frac{ I _p}{ I _s}$.
$\rightarrow$ From this equation
$\frac{v_s}{v_p}=\frac{ N _s}{N_p} \frac{ N _s}{N_p}=\frac{ I _p}{ I _s}$
$\therefore v _s=\left(\frac{ N _s}{N_p}\right) \cdot v _p \ldots(2) $
$\therefore I _s=\frac{ N _p}{N_s} \cdot I _p \cdots$
$\rightarrow$ From eq. $(2)$ and $(3)$
$(i)$ If $N _s> N _p$ then $v_s>v_p$ which means the voltage is stepped up.
Such a transformer is called step up transformer.
$(ii) $ If $N _s< N _p$ then $v _s< v _p$ which means the voltage reduces $($or voltage is stepped down$)$.
Such a transformer is called step down transformer.

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Question 43 Marks

Write the working of transformer.
###
Derive the relation $\frac{v_s}{v_p}=\frac{ N _s}{ N _p}$ for an ideal transformer and mention corresponding assumptions.

Answer

→When an AC voltage is applied to primary, the resulting current produces an alternating magnetic flux, which links the secondary and induces an emf in it. The value of this emf depends on number of turns in secondary.
→We consider an ideal transformer in which the primary has negligible resistance and all the flux in the core links both primary and secondary windings.
→Let $\phi$ be the flux in each turn in the core at time $t$ due to current in the primary when a voltage $\left(v_p\right)$ is applied to it.
→Then the induced emf or voltage (E), in the secondary
$\varepsilon_s=- N _s \frac{d \phi}{d t}$
where, $N _s$ is no. of turns in secondary.
→The alternating flux, $\phi$ also induces an emf called the back emf in the primary. This is,
$\varepsilon_p=- N _p \frac{d \phi}{d t}$
Where, $N _p$ is no. of turns in the primary.
→But $\varepsilon_p=\nu_p$ If this were not so, the primary current would be infinite due to zero resistance. (as assumed)
→If the secondary is an open circuit or the current taken from it is small, then to a good approximation, $\varepsilon_s=v_s$
→From eq. (1) and (2),
$v_s=- N _s \frac{d \phi}{d t} \ldots \text { (3) and } v_p=- N _p \frac{d \phi}{d t}$
(where $v_s$ is the voltage a cross secondary)
→By taking the ratio of eq. (3) and (4),
$\frac{v_s}{v_p}=\frac{ N _s}{N_p}$
→Following three assumptions are used in obtaining the above relation :
(1) The primary resistance and current are small;
(2) The same flux links both the primary as well as secondary (as very less flux escapes from the core)
(3) The secondary current is small.

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