101.
What is the value of inductance $$L$$ for which the current is maximum in a series $$LCR$$ circuit with $$C = 10\,\mu F$$ and $$\omega = 1000\,{s^{ - 1}}$$ ?
Current in $$LCR$$ series circuit,
$$i = \frac{V}{{\sqrt {{R^2} + {{\left( {{X_L} - {X_C}} \right)}^2}} }}$$
where, $$V$$ is $$rms$$ value of voltage $$R$$ is resistance, $${{X_L}}$$ is inductive reactance and $${{X_C}}$$ is capacitive reactance.
For current to be maximum, denominator should be minimum which can be done, i.e. during the resonance of series $$LCR$$ circuit
$$\eqalign{
& {X_L} = {X_C}\,\,{\text{i}}{\text{.e}}{\text{.}}\,\,\omega L = \frac{1}{{\omega C}} \cr
& {\text{or}}\,\,L = \frac{1}{{{\omega ^2}C}}\,\,......\left( {\text{i}} \right) \cr} $$
$$\eqalign{
& {\text{Given,}}\,\,\omega = 1000\,{s^{ - 1}},C = 10\,\mu F = 10 \times {10^{ - 6}}F \cr
& {\text{Hence,}}\,\,L = \frac{1}{{{{\left( {1000} \right)}^2} \times 10 \times {{10}^{ - 6}}}} \cr
& = 0.1\,H = 100\,mH \cr} $$
102.
A generator at a utility company produces $$100\,A$$ of current at $$4000\,V.$$ The voltage is stepped up to $$2,40,000\,V$$ by a transformer before it is sent on a high voltage transmission line. The current in transmission line is
103.
The tuning circuit of a radio receiver has a resistance of $$50\,\Omega ,$$ an inductor of $$10\,mH$$ and a variable capacitor. A $$1\,MHz$$ radio wave produces a potential difference of $$0.1\,mV.$$ The values of the capacitor to produce resonance is (Take $${\pi ^2} = 10$$ )
Key Idea
For better tuning, peak of current growth must be sharp. This is ensured by a high value of quality factor $$Q.$$
Now, quality factor is given by $$Q = \frac{1}{R}\sqrt {\frac{L}{C}} $$
From the given options, highest value of $$Q$$ is associated with $$R = 15\,\Omega ,L = 3.5\,H,C = 30\mu F$$
105.
An ideal efficient transformer has a primary power input of $$10\,kW.$$ The secondary current when the transformer is on load is $$25\,A.$$ If the primary : secondary turns ratio is $$8: 1,$$ then the potential difference applied to the primary coil is
106.
In an oscillating $$LC$$ circuit with $$L = 50\,mH$$ and $$C = 4.0\,\mu F,$$ the current is initially a maximum. How long will it take before the capacitor is fully discharged for the first time :
Time period,
$$\eqalign{
& T = 2\pi \sqrt {LC} = 2\pi \sqrt {\left( {50 \times {{10}^{ - 3}}} \right) \times 4 \times {{10}^{ - 6}}} \cr
& = 28 \times {10^{ - 4}}\,s \cr} $$
Time taken by capacitor to charge fulley,
$$t = \frac{T}{4} = 7 \times {10^{ - 4}}s.$$
107.
An $$AC$$ voltage is applied to a resistance $$R$$ and an inductor $$L$$ in series. If $$R$$ and the inductive reactance are both equal to $$3\,\Omega ,$$ the phase difference between the applied voltage and the current in the circuit is
As we know that
$$\eqalign{
& \tan \phi = \frac{{{X_L}}}{R} = \frac{{\omega L}}{R} \Rightarrow \tan \phi = \frac{3}{3} \cr
& \therefore \tan \phi = 1 \Rightarrow \phi = {45^ \circ } \cr} $$
So, phase difference $$ = \frac{\pi }{4}rad$$
108.
What is the value of inductance $$L$$ for which the current is maximum in a series $$LCR$$ circuit with $$C = 10\,\mu F$$ and $$\omega = 1000{s^{ - 1}}$$ ?
Condition for which the current is maximum in a series $$LCR$$ circuit is,
$$\eqalign{
& \omega = \frac{1}{{\sqrt {LC} }} \cr
& 1000 = \frac{1}{{\sqrt {L\left( {10 \times {{10}^{ - 6}}} \right)} }} \cr
& \Rightarrow L = 100\,mH \cr} $$
109.
A coil of inductive reactance $$31\,\Omega $$ has a resistance of $$8\,\Omega .$$ It is placed in series with a condenser of capacitative reactance $$25\,\Omega .$$ The combination is connected to an $$a.c.$$ source of $$110$$ volt. The power factor of the circuit is