Download App

Physics 5 Marks Questions

Alternating Current · Rajasthan Board (RBSE) STD 12 Science (Rajasthan - English Medium). 4 questions.

Questions

5 Marks Questions

🎯

Test yourself on this topic

4 questions · timed · auto-graded

Question 15 Marks
What is transformer ? Describe its construction and working principle.
Answer
Transformer : Transformer is a device by which an alternating voltage may be decreased or increased. This is based on the principle of mutual induction.
Construction : It consists of laminated core of soft iron, on which two coils of insulated copper wire are separately wound. These coils are kept insulated from each other and frm the iron-core, but are coupled through mutual induction. The number of turns in these coils are different. Out of these coils one coil is called primary coil and other is called the secondary coil. The terminals of primary coils are connected to AC mains and the terminals of the secondary coil are connected to external circuit in which alternating current of desired voltage is required. Transformers are of two types :
1. Step-up Transformer : It transforms the alternating low voltage to alternating high voltage and in this the number of turns in secondary coil is more than that in primary coil. (i.e., $N _s > N _p$ ).
2. Step-down Transformer : It transforms the alternating high voltage to alternating low voltage and in this the number of turns in secondary coil is less than that in primary coil (i.e., $N _s < N _p$ ).
Image
Working Principle : When alternating current source is connected to the ends of primary coil, the current changes continuously in the primary coil; due to which the magnetic flux linked with the secondary coil changes continuously, therefore the alternating emf of same frequency is developed across the secondary.
Let Np be the number of turns in primary coil, Ns the number of turns in secondary coil and I the magnetic flux linked with each turn. We assume that there is no - leakage of flux so that the flux linked with each turn of primary coil and secondary coil is the same. According to Faraday's laws the emf induced in the primary coil.
$e_p=- N _p\left(\frac{\Delta \Phi}{\Delta t}\right)$
and emf induced in the secondary coil
$e_s=- N _s\left(\frac{\Delta \Phi}{\Delta t}\right)$
$\therefore \quad \frac{e_s}{e_p}=\frac{- N _s\left(\frac{\Delta \Phi}{\Delta t}\right)}{- N _p\left(\frac{\Delta \Phi}{\Delta t}\right)} \Rightarrow \frac{e_s}{e_p}=\frac{ N _s}{N_p}$ ...(1)
If the resistance of the primary circuit is negligible then there will be no energy loss. Hence the induced emf ep in primary will be numerically equal to the applied voltage Vp across the primary. Further, if the secondary circuit is open then the voltage Vs across the terminals of the secondary coil will be equal to the induced emf es. Under these ideal conditions, we have
$\frac{e_s}{e_p}=\frac{ V _s}{V_p}$ ...(2)
Thus from equations (1) and (2), we get
$\frac{ V _s}{V_p}=\frac{ N _s}{N_p}$ ...(3)
The ratio Ns/Np = r is called the transformation ratio of the transformer. For step-up transformer Ns > Np and for step-down transformer Ns < Np hence the value of r is greater than 1 for step-up transformer and less than 1 for step-down transformer.
Hence from equation (3), we have Vs/Vp = r ⇒ Vs = r. Vp. Therefore in step-up transformer Vs > Vp and in step-down transformer Vs < Vp.
If $I_p$ and $I_s$ be the currents at any instant in primary and secondary coil and power loss be negligible, then in ideal condition :
Power in secondary = Power in primary
i.e., $V _s \times I _s= V _p \times I _p$
$\Rightarrow \quad \frac{ I _p}{ I _s}=\frac{ V _s}{V_s}$ ...(4)
$\Rightarrow \quad \frac{ I _p}{ I _s}=\frac{ N _s}{N_p}$ ...(5)
$\begin{array}{ll}\therefore & \frac{ I _p}{ I _s}=\frac{ V _s}{V_s}=\frac{ N _s}{N_p}=r\end{array}$ ...(6)
$\frac{ I _p}{ I _s}=r$
$\Rightarrow I _p=r . I _s$ and as r is greater than 1 for step-up transformer and less than 1 for step-down transformer, hence Is < Ip, in step-up transformer and Is > Ip in step down transformer.
Thus when the voltage is stepped up i.e. increased the current is correspondingly stepped down i.e.reduced in the same ratio and vice-versa under ideal conditions. But in practice it is not possible because the output power is always less than input power due to many energy losses in the transformer.
View full question & answer→
Question 25 Marks
Define mean value and root mean square value of alternating current.
Answer
Mean Value : Alternating voltage and alternating current both vary in magnitude with time and change the direction periodically. During one of the half cycles these have positive value whereas in the next half negative value. Thus the mean value of alternating voltage and alternating current each for a complete cycle is zero. Because of this an ordinary voltmeter or ammeter gives no deflection in alternating current circuit.
But for any half cycle the mean value of alternating voltage or alternating current is not zero, which can be shown as follows:
The mean value of alternating current for first positive half cycle is expressed as follows:
$I_{m}=\frac{\int_{0}^{T/2}Idt}{T/2}=\frac{\int_{0}^{T/2}(I_{0}sin \omega t)dt}{T/2}$
or $I_{m}=\frac{\int_{0}^{T/2}I_{0}sin(\frac{2\pi}{T}).tdt}
{T/2} = \frac{2I_{0}}{T}\frac{[-cos\frac{2\pi}{T}.t]_{0}^{T/2}}{\frac{2\pi}{T}}$
or $I_{m}=-\frac{2I_{0}}{T}\times\frac{T}{2\pi}[cos \pi-cos0]=-\frac{I_{0}}{\pi}[-1-1]$
or $I_{m}=+(\frac{2I_{0}}{\pi})$
or $I_{m}=+\frac{2I_{0}}{3.14}\Rightarrow I_{m}=+0.637~I_{0}$
Similarly, for next negative half cycle the average value of alternating current will be :
$I_{m}=-(\frac{2I_{0}}{\pi})$ or $I_{m}=-0.637~I_{0}$
Thus the mean value of alternating current for one complete cycle is:
$I_{m}=(+\frac{2I_{0}}{\pi})+(-\frac{2I_{0}}{\pi})=0$
Root Mean Square Value : Although the mean value of alternating voltage and alternating current both for one complete cycle is zero, the mean value of their square for one complete cycle is not zero. Therefore, rest mean square value of these is very important for alternating current circuits.
Root mean square value of alternating current is defined as follows:
"The square root of the mean value of the square of alternating current for its one complete cycle is called its root mean square value." It is denoted by $I_{rms}$ and is calculated as follows:
Mean value of $I^{2}$ for one complete cycle is given by:
$\overline{ I ^2}=\frac{\int_0^{ T } I ^2 \cdot d t}{T} \Rightarrow \overline{ I ^2}=\frac{\int_0^{ T }\left( I _0 \sin \omega t\right)^2 \cdot d t}{T}$
$\Rightarrow \quad \overline{ I ^2}=\frac{\int_0^{ T } I _0^2 \sin ^2 \omega t \cdot d t}{T}=\frac{\int_0^{ T } I _0^2 \sin ^2\left(\frac{2 \pi}{T}\right) \cdot t \cdot d t}{T}$
$\Rightarrow \quad \overline{ I ^2}=\frac{\int_0^{ T } I _0^2 2 \sin ^2\left(\frac{2 \pi}{T}\right) \cdot t \cdot d t}{2 T}$
or $\overline{ I ^2}=\frac{ I _0^2}{2 T} \int_0^{ T }\left[1-\cos 2 \times\left(\frac{2 \pi}{T}\right) \cdot t\right] d t$
$\Rightarrow \quad \overline{ I ^2}=\frac{ I _0^2}{2 T}\left[\int_0^{ T } d t-\int_0^{ T } \cos \left(\frac{4 \pi}{T}\right) \cdot t \cdot d t\right]$
$\Rightarrow \quad \overline{ I ^2}=\frac{ I _0^2}{2 T}\left[\frac{\{t\}_0^{ T }-\left[\sin \left(\frac{4 \pi}{T}\right) \cdot t\right]_0^{ T }}{\frac{4 \pi}{T}}\right]$
$\Rightarrow \quad \overline{ I ^2}=\frac{ I _0^2}{2 T}\left[\{ T \}-\left\{\frac{ T }{4 \pi}\left(\sin 4 \pi-\sin 0^0\right)\right\}\right]$
$\Rightarrow \quad \overline{ I ^2}=\frac{ I _0^2}{2 T} \times T -\frac{ I _0^2}{2 T} \times 0 \Rightarrow \overline{ I ^2}=\frac{ I _0^2}{2}$
But according to the definition of root mean square value of alternating current it is given by :
$I _{r m s}=\sqrt{\overline{\overline{ I ^2}}} \Rightarrow I _{r m s}=\sqrt{\left(\frac{ I _0^2}{2}\right)}$
or $I _{r m s}=\frac{ I _0}{\sqrt{2}}$
or $I_{r m s}=\frac{I_0}{1.414} \Rightarrow I_{r m s}=0.707 I_0$
View full question & answer→
Question 35 Marks
Draw the graph showing the variation of alternating current with frequency in a L-C-R series resonant circuit and derive the expression for band width.
Answer
Variation of current with frequency (an-gular frequency) in a series L-C-R resonant cir-cuit is shown in adjoining figure.
It is clear that for frequencies greater than or less than $\omega_0$, the values of current are less than the maximum value (I), where $\omega_0$ is resonant angular frequency.
Expression for Band Width : In series LCR resonant circuit at resonant angular frequency $\omega_0=1 / \sqrt{ LC }$ the amplitude of current is maximum. The quickness with which the current falls from its resonant value $\left( V _{ o } /\right.$R) with change in applied frequency is known as sharpness of resonance.
"It is measured by the ratio of resonant frequency $\omega_0$ to the diffe-rence of two frequencies $\omega_1$ and $\omega_2$ at which the current falls to $1 / \sqrt{2}$ of the resonant value.
Image
∴ Sharpness of resonance $=\frac{\omega_0}{\omega_2-\omega_1}$ Here $\omega_1$ and $\omega_2$ known as half power frequencies, because at these frequencies the power in the circuit reduces to half of its maximum value. The difference of half power frequencies ($\omega_1$ - $\omega_2$) is known as 'band width'. The smaller is the band width, the sharper is the resonance.
Expression for Band Width : The value of impedance of series LCR circuit at resonant frequency is Z, therefore at $\omega_1$ and $\omega_2$ it will be $\sqrt{2} R$ because at these frequencies current falls to $1 / \sqrt{2}$ of the resonant value.
$\therefore Z=\sqrt{2} R \Rightarrow \sqrt{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2}=\sqrt{2} \cdot R$
On squaring both sides,
$R ^2+\left(\omega L -\frac{1}{\omega C }\right)^2=2 R ^2$
or $\left(\omega L -\frac{1}{\omega C }\right)^2= R ^2$
or $\left(\omega L -\frac{1}{\omega C }\right)= \pm R$
Thus, if $\omega_2$ > $\omega_1$ then it may be concluded that
$\omega_1 L-\frac{1}{\omega_1 C}=-R$ ...(1)
and $\omega_2 L-\frac{1}{\omega_2 C}=+R$ ...(2)
On adding equations (1) and (2) get
$\left(\omega_1-\omega_2\right) L-\frac{1}{C}\left(\frac{\omega_1-\omega_2}{\omega_1 \omega_2}\right)=0$
$\Rightarrow \quad \omega_2 \omega_1=\frac{1}{ LC }$ ...(3)
On subtracting eq. (1) from (2), we get
$\left(\omega_2-\omega_1\right) L+\frac{1}{C}\left(\frac{\omega_2-\omega_1}{\omega_1 \omega_2}\right)=2 R$
or $\left(\omega_2-\omega_1\right)\left[L+\frac{1}{C \omega_1 \omega_2}\right]=2 R$ ...(4)
On substituting the value of $\omega_1 \omega_2$ from equation (3) in equation (4),
$\left(\omega_2-\omega_1\right)\left[ L +\frac{1}{ C \times \frac{1}{ LC }}\right]=2 R$
$\Rightarrow \quad\left(\omega_2-\omega_1\right)= R / L$
i.e., Band width $=( R / L )$
View full question & answer→
Question 45 Marks
In an A.C. circuit inductance L, capacitance C and resistance R are connected in series. Find the expression for impedance Z of circuit and phase difference between V and I.### Prove that : $Z=\sqrt{R^{2}+(\omega L-\frac{1}{\omega C})^{2}}$ and $tan \phi=\frac{(\omega L-\frac{1}{\omega C})}{R}$
Answer
Circuit diagram of L-C-R series A.C. circuit is shown in the following fig. (a)
Image
Potential difference $V_{R}$ across resistance R and current I in the resistance are is same phase. Induced voltage $V_{L}$ across inductance L will lead current by $90^{\circ}$ and voltage $V_{c}$ across plates of capacitor will lag the current by $90^{\circ}$. Hence phase difference between $V_{L}$ and $V_{c}$ will be $180^{\circ}$ and resultant potential of $V_{L}$ and $V_{c}$ will be $V_{L}-V_{C}$ (when $V_{L}>V_{C})$.
If V be the potential of L-C-R circuit, then
$V^{2}=V_{R}^{2}+(V_{L}-V_{C})^{2}$
But $V_{R}=IR, V_{L}=I \times X_{L}$ and $V_{C}=I X_{C}$
$\therefore$ $V^{2}=I^{2}R^{2}+(IX_{L}-IX_{C})^{2}
= I^{2}[R^{2}+(X_{L}-X_{C})^{2}]$
$V/I=\sqrt{\{R^{2}+(X_{L}-X_{C})^{2}\}}$
V/I is called impedance of the circuit and is denoted by Z.
$\therefore$ $Z=\sqrt{\{R^{2}+(\omega L-\frac{1}{\omega C})^{2}\}}$
But $X_L=\omega L$ and $X_C=1 / \omega C$
\[\text { ∴ Impedance } Z=\sqrt{\left\{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2\right\}}\] ...(1)
If $\phi$ be the phase difference between voltage V and current I from fig (b) ,
$\tan \phi=\frac{V_L-V_C}{V_R}=\frac{I\left(X_L-X_C\right)}{I R}$
\[\tan \phi=\frac{\left(X_L-X_C\right)}{R} \Rightarrow \tan \theta=\frac{\left(\omega L-\frac{1}{\omega C}\right)}{R} \ldots\] ...(2)
\[\text { ⇒ Phase difference } \phi= \operatorname { t a n } ^{-1}\left[\frac{\left(\omega L -\frac{ 1 }{\omega C }\right)}{ R }\right] \ldots\] ...(3)
Thus phase $\phi$ depends upon $R$ and relative values of $X_L$ and $X_{C^{\prime}}$.
View full question & answer→
Generate Paper FreeAll Alternating Current topics