155060
A bulb and a condenser are connected in series with an A.C. source. On increasing the frequency of the source its brightness will
1 Increase
2 Decrease
3 Sometimes increase and sometimes decrease
4 Neither increase nor decrease.
Explanation:
A Capacitive reactance and frequency has an inverse dependent $\mathrm{X}_{\mathrm{C}} \propto \frac{1}{\mathrm{f}}$ Thus, frequency of an A.C source increases, the capacitive reactance decreases. Therefore, the bulb glow with more brightness because more current flows through it.
COMEDK 2018
Alternating Current
155061
In a series $L R$ circuit $(L=3 H, R=1.5 \Omega)$ and DC voltage $=1 \mathrm{~V}$. Find current at $T=2$ seconds.
1 \(0.4 \mathrm{~A}\)
2 \(0.6 \mathrm{~A}\)
3 \(0.8 \mathrm{~A}\)
4 \(0.9 \mathrm{~A}\)
Explanation:
A $\mathrm{L}=3 \mathrm{H}, \mathrm{R}=1.5 \Omega, \mathrm{E}=1 \mathrm{~V}, \mathrm{~T}=2 \mathrm{sec}$ For LR circuit, $\tau=\frac{\mathrm{L}}{\mathrm{R}}=\frac{3}{1.5}=2$ $\mathrm{I}=\frac{\mathrm{E}}{\mathrm{R}}\left[1-\mathrm{e}^{-\mathrm{t} / \tau}\right]=\frac{1}{1.5}\left[1-\mathrm{e}^{-2 / 2}\right]$ $\mathrm{I}=\frac{2}{3}\left[1-\frac{1}{\mathrm{e}}\right]=\frac{2}{3}(1-0.367)$ $\sqcup 0.4 \mathrm{~A}$
AIIMS-26.05.2018(E)
Alternating Current
155065
Time constant of a series $\mathrm{R}-\mathrm{C}$ circuit is
1 $+\mathrm{RC}$
2 $-\mathrm{RC}$
3 $\mathrm{R} / \mathrm{C}$
4 $\mathrm{C} / \mathrm{R}$
Explanation:
A We know that the growth of current in RC circuit is given by $\mathrm{i}=\mathrm{i}_{\max } \mathrm{e}^{-\mathrm{t} / \mathrm{RC}}$ Where $\mathrm{RC}=$ Product of resistance and capacitance in circuit which is known as time constant and its unit is second.
CG PET- 2014
Alternating Current
155073
The quantities $R C$ and $\left(\frac{L}{R}\right)$ (where $R, L$ and $C$ stand for resistance, inductance and capacitance respectively) have the dimensions of
1 force
2 linear momentum
3 linear acceleration
4 time
Explanation:
D $\text { } : \mathrm{RC}=\text { ohm } \times \text { farad }$ $=\text { ohm } \times \text { coulomb } / \text { volt }$ $=\frac{\text { Volt }}{\text { Amp }} \times \frac{\text { Coulomb }}{\text { Volt }}=\frac{\text { Coulomb }}{\text { Ampere }}=\text { Second }[\mathrm{T}]$ $\frac{\mathrm{L}}{\mathrm{R}}=\frac{\text { henery }}{\mathrm{ohm}}=\frac{\text { ohm } \times \text { second }}{\text { ohm }}=\text { Second }[\mathrm{T}]$ $\text { Both RC and } \frac{\mathrm{L}}{\mathrm{R}} \text { have the dimension of time }$
155060
A bulb and a condenser are connected in series with an A.C. source. On increasing the frequency of the source its brightness will
1 Increase
2 Decrease
3 Sometimes increase and sometimes decrease
4 Neither increase nor decrease.
Explanation:
A Capacitive reactance and frequency has an inverse dependent $\mathrm{X}_{\mathrm{C}} \propto \frac{1}{\mathrm{f}}$ Thus, frequency of an A.C source increases, the capacitive reactance decreases. Therefore, the bulb glow with more brightness because more current flows through it.
COMEDK 2018
Alternating Current
155061
In a series $L R$ circuit $(L=3 H, R=1.5 \Omega)$ and DC voltage $=1 \mathrm{~V}$. Find current at $T=2$ seconds.
1 \(0.4 \mathrm{~A}\)
2 \(0.6 \mathrm{~A}\)
3 \(0.8 \mathrm{~A}\)
4 \(0.9 \mathrm{~A}\)
Explanation:
A $\mathrm{L}=3 \mathrm{H}, \mathrm{R}=1.5 \Omega, \mathrm{E}=1 \mathrm{~V}, \mathrm{~T}=2 \mathrm{sec}$ For LR circuit, $\tau=\frac{\mathrm{L}}{\mathrm{R}}=\frac{3}{1.5}=2$ $\mathrm{I}=\frac{\mathrm{E}}{\mathrm{R}}\left[1-\mathrm{e}^{-\mathrm{t} / \tau}\right]=\frac{1}{1.5}\left[1-\mathrm{e}^{-2 / 2}\right]$ $\mathrm{I}=\frac{2}{3}\left[1-\frac{1}{\mathrm{e}}\right]=\frac{2}{3}(1-0.367)$ $\sqcup 0.4 \mathrm{~A}$
AIIMS-26.05.2018(E)
Alternating Current
155065
Time constant of a series $\mathrm{R}-\mathrm{C}$ circuit is
1 $+\mathrm{RC}$
2 $-\mathrm{RC}$
3 $\mathrm{R} / \mathrm{C}$
4 $\mathrm{C} / \mathrm{R}$
Explanation:
A We know that the growth of current in RC circuit is given by $\mathrm{i}=\mathrm{i}_{\max } \mathrm{e}^{-\mathrm{t} / \mathrm{RC}}$ Where $\mathrm{RC}=$ Product of resistance and capacitance in circuit which is known as time constant and its unit is second.
CG PET- 2014
Alternating Current
155073
The quantities $R C$ and $\left(\frac{L}{R}\right)$ (where $R, L$ and $C$ stand for resistance, inductance and capacitance respectively) have the dimensions of
1 force
2 linear momentum
3 linear acceleration
4 time
Explanation:
D $\text { } : \mathrm{RC}=\text { ohm } \times \text { farad }$ $=\text { ohm } \times \text { coulomb } / \text { volt }$ $=\frac{\text { Volt }}{\text { Amp }} \times \frac{\text { Coulomb }}{\text { Volt }}=\frac{\text { Coulomb }}{\text { Ampere }}=\text { Second }[\mathrm{T}]$ $\frac{\mathrm{L}}{\mathrm{R}}=\frac{\text { henery }}{\mathrm{ohm}}=\frac{\text { ohm } \times \text { second }}{\text { ohm }}=\text { Second }[\mathrm{T}]$ $\text { Both RC and } \frac{\mathrm{L}}{\mathrm{R}} \text { have the dimension of time }$
155060
A bulb and a condenser are connected in series with an A.C. source. On increasing the frequency of the source its brightness will
1 Increase
2 Decrease
3 Sometimes increase and sometimes decrease
4 Neither increase nor decrease.
Explanation:
A Capacitive reactance and frequency has an inverse dependent $\mathrm{X}_{\mathrm{C}} \propto \frac{1}{\mathrm{f}}$ Thus, frequency of an A.C source increases, the capacitive reactance decreases. Therefore, the bulb glow with more brightness because more current flows through it.
COMEDK 2018
Alternating Current
155061
In a series $L R$ circuit $(L=3 H, R=1.5 \Omega)$ and DC voltage $=1 \mathrm{~V}$. Find current at $T=2$ seconds.
1 \(0.4 \mathrm{~A}\)
2 \(0.6 \mathrm{~A}\)
3 \(0.8 \mathrm{~A}\)
4 \(0.9 \mathrm{~A}\)
Explanation:
A $\mathrm{L}=3 \mathrm{H}, \mathrm{R}=1.5 \Omega, \mathrm{E}=1 \mathrm{~V}, \mathrm{~T}=2 \mathrm{sec}$ For LR circuit, $\tau=\frac{\mathrm{L}}{\mathrm{R}}=\frac{3}{1.5}=2$ $\mathrm{I}=\frac{\mathrm{E}}{\mathrm{R}}\left[1-\mathrm{e}^{-\mathrm{t} / \tau}\right]=\frac{1}{1.5}\left[1-\mathrm{e}^{-2 / 2}\right]$ $\mathrm{I}=\frac{2}{3}\left[1-\frac{1}{\mathrm{e}}\right]=\frac{2}{3}(1-0.367)$ $\sqcup 0.4 \mathrm{~A}$
AIIMS-26.05.2018(E)
Alternating Current
155065
Time constant of a series $\mathrm{R}-\mathrm{C}$ circuit is
1 $+\mathrm{RC}$
2 $-\mathrm{RC}$
3 $\mathrm{R} / \mathrm{C}$
4 $\mathrm{C} / \mathrm{R}$
Explanation:
A We know that the growth of current in RC circuit is given by $\mathrm{i}=\mathrm{i}_{\max } \mathrm{e}^{-\mathrm{t} / \mathrm{RC}}$ Where $\mathrm{RC}=$ Product of resistance and capacitance in circuit which is known as time constant and its unit is second.
CG PET- 2014
Alternating Current
155073
The quantities $R C$ and $\left(\frac{L}{R}\right)$ (where $R, L$ and $C$ stand for resistance, inductance and capacitance respectively) have the dimensions of
1 force
2 linear momentum
3 linear acceleration
4 time
Explanation:
D $\text { } : \mathrm{RC}=\text { ohm } \times \text { farad }$ $=\text { ohm } \times \text { coulomb } / \text { volt }$ $=\frac{\text { Volt }}{\text { Amp }} \times \frac{\text { Coulomb }}{\text { Volt }}=\frac{\text { Coulomb }}{\text { Ampere }}=\text { Second }[\mathrm{T}]$ $\frac{\mathrm{L}}{\mathrm{R}}=\frac{\text { henery }}{\mathrm{ohm}}=\frac{\text { ohm } \times \text { second }}{\text { ohm }}=\text { Second }[\mathrm{T}]$ $\text { Both RC and } \frac{\mathrm{L}}{\mathrm{R}} \text { have the dimension of time }$
155060
A bulb and a condenser are connected in series with an A.C. source. On increasing the frequency of the source its brightness will
1 Increase
2 Decrease
3 Sometimes increase and sometimes decrease
4 Neither increase nor decrease.
Explanation:
A Capacitive reactance and frequency has an inverse dependent $\mathrm{X}_{\mathrm{C}} \propto \frac{1}{\mathrm{f}}$ Thus, frequency of an A.C source increases, the capacitive reactance decreases. Therefore, the bulb glow with more brightness because more current flows through it.
COMEDK 2018
Alternating Current
155061
In a series $L R$ circuit $(L=3 H, R=1.5 \Omega)$ and DC voltage $=1 \mathrm{~V}$. Find current at $T=2$ seconds.
1 \(0.4 \mathrm{~A}\)
2 \(0.6 \mathrm{~A}\)
3 \(0.8 \mathrm{~A}\)
4 \(0.9 \mathrm{~A}\)
Explanation:
A $\mathrm{L}=3 \mathrm{H}, \mathrm{R}=1.5 \Omega, \mathrm{E}=1 \mathrm{~V}, \mathrm{~T}=2 \mathrm{sec}$ For LR circuit, $\tau=\frac{\mathrm{L}}{\mathrm{R}}=\frac{3}{1.5}=2$ $\mathrm{I}=\frac{\mathrm{E}}{\mathrm{R}}\left[1-\mathrm{e}^{-\mathrm{t} / \tau}\right]=\frac{1}{1.5}\left[1-\mathrm{e}^{-2 / 2}\right]$ $\mathrm{I}=\frac{2}{3}\left[1-\frac{1}{\mathrm{e}}\right]=\frac{2}{3}(1-0.367)$ $\sqcup 0.4 \mathrm{~A}$
AIIMS-26.05.2018(E)
Alternating Current
155065
Time constant of a series $\mathrm{R}-\mathrm{C}$ circuit is
1 $+\mathrm{RC}$
2 $-\mathrm{RC}$
3 $\mathrm{R} / \mathrm{C}$
4 $\mathrm{C} / \mathrm{R}$
Explanation:
A We know that the growth of current in RC circuit is given by $\mathrm{i}=\mathrm{i}_{\max } \mathrm{e}^{-\mathrm{t} / \mathrm{RC}}$ Where $\mathrm{RC}=$ Product of resistance and capacitance in circuit which is known as time constant and its unit is second.
CG PET- 2014
Alternating Current
155073
The quantities $R C$ and $\left(\frac{L}{R}\right)$ (where $R, L$ and $C$ stand for resistance, inductance and capacitance respectively) have the dimensions of
1 force
2 linear momentum
3 linear acceleration
4 time
Explanation:
D $\text { } : \mathrm{RC}=\text { ohm } \times \text { farad }$ $=\text { ohm } \times \text { coulomb } / \text { volt }$ $=\frac{\text { Volt }}{\text { Amp }} \times \frac{\text { Coulomb }}{\text { Volt }}=\frac{\text { Coulomb }}{\text { Ampere }}=\text { Second }[\mathrm{T}]$ $\frac{\mathrm{L}}{\mathrm{R}}=\frac{\text { henery }}{\mathrm{ohm}}=\frac{\text { ohm } \times \text { second }}{\text { ohm }}=\text { Second }[\mathrm{T}]$ $\text { Both RC and } \frac{\mathrm{L}}{\mathrm{R}} \text { have the dimension of time }$
