155190
In a circuit $L, C$ and $R$ are connected in series with an alternating voltage source of frequency f. The current leads the voltage by $45^{\circ}$. The value of $C$ is
C Given that, Phase difference between current and voltage of an series L-C-R circuit is $=45^{\circ}$ We know that, $\tan \theta=\frac{X_{L}-X_{C}}{R}$ $\tan 45^{\circ}=\frac{X_{L}-X_{C}}{R}$ $1=\frac{X_{L}-X_{C}}{R}$ $R=X_{L}-X_{C}$ $=2 \pi f L-\frac{1}{2 \pi f C}$ $\frac{1}{2 \pi f C}=2 \pi f L-R$ $C=\frac{1}{2 \pi f(2 \pi f L-R)}$
Manipal UGET-2010
Alternating Current
155191
In series $L R$ circuit $X_{L}=3 R$, now a capacitor with $X_{C}=R$ is added in series. The ratio of new to old power factor is
1 1
2 2
3 $\frac{1}{\sqrt{2}}$
4 $\sqrt{2}$
Explanation:
D Given, $X_{L}=3 R, X_{C}=R$ Condition 1 - power factor for $\mathrm{R}-\mathrm{C}$ circuit is - $(\cos \phi)_{\text {old }}=\frac{\mathrm{R}}{\mathrm{Z}}$ Impedance, $\mathrm{Z}=\sqrt{\mathrm{X}_{\mathrm{L}}^{2}+\mathrm{R}^{2}}$ $Z=\sqrt{(3 R)^{2}+R^{2}}$ $Z=\sqrt{10} R$ $(\cos \phi)_{\text {old }}=\frac{R}{\sqrt{10} R}=\frac{1}{\sqrt{10}}$ Condition 2 - If $X_{C}=R$ is added then Impedance, $Z_{\text {new }}=\sqrt{\left(X_{L}-X_{C}\right)^{2}+R^{2}}$ $=\sqrt{R^{2}+(3 R-R)^{2}}=\sqrt{5} R$ New power factor, $(\cos \phi)_{\text {new }}=\frac{\mathrm{R}}{\sqrt{5} \mathrm{R}}=\frac{1}{\sqrt{5}}$ $\frac{(\cos \phi)_{\text {new }}}{(\cos \phi)_{\text {old }}}=\frac{1 / \sqrt{5}}{1 / \sqrt{10}}=\sqrt{\frac{10}{5}}=\sqrt{2}$ The ratio of new to old power factor is $\sqrt{2}$.
CG PET -2016
Alternating Current
155192
If $\mathrm{L}, \mathrm{C}, \mathrm{R}$ are respectively the inductance, capacitance and resistance, the quantities of dimensions same as of frequency are
1 $\frac{1}{\sqrt{\mathrm{LC}}}, \frac{\mathrm{R}}{\mathrm{L}}$ and $\frac{1}{\mathrm{RC}}$
2 $\sqrt{\mathrm{LC}}, \frac{\mathrm{L}}{\mathrm{R}}$ and $\mathrm{RC}$
3 $\sqrt{\frac{\mathrm{L}}{\mathrm{C}}}, \mathrm{LR}$ and $\frac{\mathrm{C}}{\mathrm{R}}$
4 $\sqrt{\frac{\mathrm{C}}{\mathrm{L}}}, \frac{1}{\mathrm{LR}}$ and $\frac{\mathrm{R}}{\mathrm{C}}$
Explanation:
A Dimension of $\frac{1}{\sqrt{\mathrm{LC}}}$ $=\frac{1}{\left[\mathrm{ML}^{2} \mathrm{~T}^{-2} \mathrm{~A}^{-2}\right]^{1 / 2}\left[\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^{4} \mathrm{~A}^{2}\right]^{1 / 2}}$ $=\frac{1}{[\mathrm{~T}]}=\left[\mathrm{T}^{-1}\right]$ Dimension of $\frac{\mathrm{R}}{\mathrm{L}}=\frac{\left[\mathrm{ML}^{2} \mathrm{~T}^{-3} \mathrm{~A}^{-2}\right]}{\left[\mathrm{ML}^{2} \mathrm{~T}^{-2} \mathrm{~A}^{-2}\right]}=\left[\mathrm{T}^{-1}\right]$ Dimension of $\frac{1}{\mathrm{RC}}=\frac{1}{\left[\mathrm{ML}^{2} \mathrm{~T}^{-3} \mathrm{~A}^{-2}\right]\left[\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^{4} \mathrm{~A}^{2}\right]}$ $=\left[\mathrm{T}^{-1}\right]$ Dimension of frequency $=\left[\mathrm{T}^{-1}\right]$ Hence, $\frac{1}{\sqrt{\mathrm{LC}}}, \frac{\mathrm{R}}{\mathrm{L}}$ and $\frac{1}{\mathrm{RC}}$ have the same dimension as frequency.
CG PET- 2011
Alternating Current
155193
$L-C-R$ circuit $V_{L}=V_{C}=V_{R}=10 V$ if $C$ is short circuited then voltage across $\mathrm{L}$ will be
1 $10 / \sqrt{2} \mathrm{~V}$
2 $20 \sqrt{2} \mathrm{~V}$
3 $10 \sqrt{2} \mathrm{~V}$
4 $10 \mathrm{~V}$
Explanation:
A Given, $V_{L}=V_{C}=V_{R}=10 \mathrm{~V}$ Then source voltage is given by - $\mathrm{V}=\sqrt{\mathrm{V}_{\mathrm{R}}^{2}+\left(\mathrm{V}_{\mathrm{L}}-\mathrm{V}_{\mathrm{C}}\right)^{2}}$ $\mathrm{~V}=\sqrt{(10)^{2}+(10-10)^{2}}$ $\mathrm{~V}=10 \mathrm{~V}$ Now if $\mathrm{C}$ is short circuited then voltage across capacitor is zero $\mathrm{V}_{\mathrm{C}}=0$ $\because \quad \mathrm{V}_{\mathrm{L}}, \mathrm{V}_{\mathrm{C}}$ and $\mathrm{V}_{\mathrm{R}}$ are equal $\text { And } \quad I=\frac{V}{\sqrt{R^{2}+X_{L}^{2}}}=\frac{10}{\sqrt{R^{2}+R^{2}}}=\frac{10}{\sqrt{2} R}$ Voltage across inductor is - $\mathrm{V}_{\mathrm{L}}=\mathrm{IX}_{\mathrm{L}}=\frac{10}{\sqrt{2} \mathrm{R}} \cdot \mathrm{R} \quad\left\{\because \mathrm{X}_{\mathrm{L}}=\mathrm{R}\right\}$ $\mathrm{V}_{\mathrm{L}}=\frac{10}{\sqrt{2}} \mathrm{~V}$
CG PET- 2007
Alternating Current
155194
The impedance of an $\mathrm{AC}$ circuit containing a capacitive reactance of $5 \Omega$ and inductive reactance of $8 \Omega$ will be
1 $1.6 \Omega$
2 $40 \Omega$
3 $3 \Omega$
4 $13 \Omega$
Explanation:
C Given that, Capacitive reactance $\mathrm{X}_{\mathrm{C}}=5 \Omega$, Inductive reactance $\mathrm{X}_{\mathrm{L}}=8 \Omega$, Impedance of $\mathrm{LC}$ circuits $Z=\sqrt{\left(X_{L}-X_{C}\right)^{2}}=\sqrt{(8-5)^{2}}=\sqrt{9}=3$ $Z=3 \Omega$
155190
In a circuit $L, C$ and $R$ are connected in series with an alternating voltage source of frequency f. The current leads the voltage by $45^{\circ}$. The value of $C$ is
C Given that, Phase difference between current and voltage of an series L-C-R circuit is $=45^{\circ}$ We know that, $\tan \theta=\frac{X_{L}-X_{C}}{R}$ $\tan 45^{\circ}=\frac{X_{L}-X_{C}}{R}$ $1=\frac{X_{L}-X_{C}}{R}$ $R=X_{L}-X_{C}$ $=2 \pi f L-\frac{1}{2 \pi f C}$ $\frac{1}{2 \pi f C}=2 \pi f L-R$ $C=\frac{1}{2 \pi f(2 \pi f L-R)}$
Manipal UGET-2010
Alternating Current
155191
In series $L R$ circuit $X_{L}=3 R$, now a capacitor with $X_{C}=R$ is added in series. The ratio of new to old power factor is
1 1
2 2
3 $\frac{1}{\sqrt{2}}$
4 $\sqrt{2}$
Explanation:
D Given, $X_{L}=3 R, X_{C}=R$ Condition 1 - power factor for $\mathrm{R}-\mathrm{C}$ circuit is - $(\cos \phi)_{\text {old }}=\frac{\mathrm{R}}{\mathrm{Z}}$ Impedance, $\mathrm{Z}=\sqrt{\mathrm{X}_{\mathrm{L}}^{2}+\mathrm{R}^{2}}$ $Z=\sqrt{(3 R)^{2}+R^{2}}$ $Z=\sqrt{10} R$ $(\cos \phi)_{\text {old }}=\frac{R}{\sqrt{10} R}=\frac{1}{\sqrt{10}}$ Condition 2 - If $X_{C}=R$ is added then Impedance, $Z_{\text {new }}=\sqrt{\left(X_{L}-X_{C}\right)^{2}+R^{2}}$ $=\sqrt{R^{2}+(3 R-R)^{2}}=\sqrt{5} R$ New power factor, $(\cos \phi)_{\text {new }}=\frac{\mathrm{R}}{\sqrt{5} \mathrm{R}}=\frac{1}{\sqrt{5}}$ $\frac{(\cos \phi)_{\text {new }}}{(\cos \phi)_{\text {old }}}=\frac{1 / \sqrt{5}}{1 / \sqrt{10}}=\sqrt{\frac{10}{5}}=\sqrt{2}$ The ratio of new to old power factor is $\sqrt{2}$.
CG PET -2016
Alternating Current
155192
If $\mathrm{L}, \mathrm{C}, \mathrm{R}$ are respectively the inductance, capacitance and resistance, the quantities of dimensions same as of frequency are
1 $\frac{1}{\sqrt{\mathrm{LC}}}, \frac{\mathrm{R}}{\mathrm{L}}$ and $\frac{1}{\mathrm{RC}}$
2 $\sqrt{\mathrm{LC}}, \frac{\mathrm{L}}{\mathrm{R}}$ and $\mathrm{RC}$
3 $\sqrt{\frac{\mathrm{L}}{\mathrm{C}}}, \mathrm{LR}$ and $\frac{\mathrm{C}}{\mathrm{R}}$
4 $\sqrt{\frac{\mathrm{C}}{\mathrm{L}}}, \frac{1}{\mathrm{LR}}$ and $\frac{\mathrm{R}}{\mathrm{C}}$
Explanation:
A Dimension of $\frac{1}{\sqrt{\mathrm{LC}}}$ $=\frac{1}{\left[\mathrm{ML}^{2} \mathrm{~T}^{-2} \mathrm{~A}^{-2}\right]^{1 / 2}\left[\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^{4} \mathrm{~A}^{2}\right]^{1 / 2}}$ $=\frac{1}{[\mathrm{~T}]}=\left[\mathrm{T}^{-1}\right]$ Dimension of $\frac{\mathrm{R}}{\mathrm{L}}=\frac{\left[\mathrm{ML}^{2} \mathrm{~T}^{-3} \mathrm{~A}^{-2}\right]}{\left[\mathrm{ML}^{2} \mathrm{~T}^{-2} \mathrm{~A}^{-2}\right]}=\left[\mathrm{T}^{-1}\right]$ Dimension of $\frac{1}{\mathrm{RC}}=\frac{1}{\left[\mathrm{ML}^{2} \mathrm{~T}^{-3} \mathrm{~A}^{-2}\right]\left[\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^{4} \mathrm{~A}^{2}\right]}$ $=\left[\mathrm{T}^{-1}\right]$ Dimension of frequency $=\left[\mathrm{T}^{-1}\right]$ Hence, $\frac{1}{\sqrt{\mathrm{LC}}}, \frac{\mathrm{R}}{\mathrm{L}}$ and $\frac{1}{\mathrm{RC}}$ have the same dimension as frequency.
CG PET- 2011
Alternating Current
155193
$L-C-R$ circuit $V_{L}=V_{C}=V_{R}=10 V$ if $C$ is short circuited then voltage across $\mathrm{L}$ will be
1 $10 / \sqrt{2} \mathrm{~V}$
2 $20 \sqrt{2} \mathrm{~V}$
3 $10 \sqrt{2} \mathrm{~V}$
4 $10 \mathrm{~V}$
Explanation:
A Given, $V_{L}=V_{C}=V_{R}=10 \mathrm{~V}$ Then source voltage is given by - $\mathrm{V}=\sqrt{\mathrm{V}_{\mathrm{R}}^{2}+\left(\mathrm{V}_{\mathrm{L}}-\mathrm{V}_{\mathrm{C}}\right)^{2}}$ $\mathrm{~V}=\sqrt{(10)^{2}+(10-10)^{2}}$ $\mathrm{~V}=10 \mathrm{~V}$ Now if $\mathrm{C}$ is short circuited then voltage across capacitor is zero $\mathrm{V}_{\mathrm{C}}=0$ $\because \quad \mathrm{V}_{\mathrm{L}}, \mathrm{V}_{\mathrm{C}}$ and $\mathrm{V}_{\mathrm{R}}$ are equal $\text { And } \quad I=\frac{V}{\sqrt{R^{2}+X_{L}^{2}}}=\frac{10}{\sqrt{R^{2}+R^{2}}}=\frac{10}{\sqrt{2} R}$ Voltage across inductor is - $\mathrm{V}_{\mathrm{L}}=\mathrm{IX}_{\mathrm{L}}=\frac{10}{\sqrt{2} \mathrm{R}} \cdot \mathrm{R} \quad\left\{\because \mathrm{X}_{\mathrm{L}}=\mathrm{R}\right\}$ $\mathrm{V}_{\mathrm{L}}=\frac{10}{\sqrt{2}} \mathrm{~V}$
CG PET- 2007
Alternating Current
155194
The impedance of an $\mathrm{AC}$ circuit containing a capacitive reactance of $5 \Omega$ and inductive reactance of $8 \Omega$ will be
1 $1.6 \Omega$
2 $40 \Omega$
3 $3 \Omega$
4 $13 \Omega$
Explanation:
C Given that, Capacitive reactance $\mathrm{X}_{\mathrm{C}}=5 \Omega$, Inductive reactance $\mathrm{X}_{\mathrm{L}}=8 \Omega$, Impedance of $\mathrm{LC}$ circuits $Z=\sqrt{\left(X_{L}-X_{C}\right)^{2}}=\sqrt{(8-5)^{2}}=\sqrt{9}=3$ $Z=3 \Omega$
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Alternating Current
155190
In a circuit $L, C$ and $R$ are connected in series with an alternating voltage source of frequency f. The current leads the voltage by $45^{\circ}$. The value of $C$ is
C Given that, Phase difference between current and voltage of an series L-C-R circuit is $=45^{\circ}$ We know that, $\tan \theta=\frac{X_{L}-X_{C}}{R}$ $\tan 45^{\circ}=\frac{X_{L}-X_{C}}{R}$ $1=\frac{X_{L}-X_{C}}{R}$ $R=X_{L}-X_{C}$ $=2 \pi f L-\frac{1}{2 \pi f C}$ $\frac{1}{2 \pi f C}=2 \pi f L-R$ $C=\frac{1}{2 \pi f(2 \pi f L-R)}$
Manipal UGET-2010
Alternating Current
155191
In series $L R$ circuit $X_{L}=3 R$, now a capacitor with $X_{C}=R$ is added in series. The ratio of new to old power factor is
1 1
2 2
3 $\frac{1}{\sqrt{2}}$
4 $\sqrt{2}$
Explanation:
D Given, $X_{L}=3 R, X_{C}=R$ Condition 1 - power factor for $\mathrm{R}-\mathrm{C}$ circuit is - $(\cos \phi)_{\text {old }}=\frac{\mathrm{R}}{\mathrm{Z}}$ Impedance, $\mathrm{Z}=\sqrt{\mathrm{X}_{\mathrm{L}}^{2}+\mathrm{R}^{2}}$ $Z=\sqrt{(3 R)^{2}+R^{2}}$ $Z=\sqrt{10} R$ $(\cos \phi)_{\text {old }}=\frac{R}{\sqrt{10} R}=\frac{1}{\sqrt{10}}$ Condition 2 - If $X_{C}=R$ is added then Impedance, $Z_{\text {new }}=\sqrt{\left(X_{L}-X_{C}\right)^{2}+R^{2}}$ $=\sqrt{R^{2}+(3 R-R)^{2}}=\sqrt{5} R$ New power factor, $(\cos \phi)_{\text {new }}=\frac{\mathrm{R}}{\sqrt{5} \mathrm{R}}=\frac{1}{\sqrt{5}}$ $\frac{(\cos \phi)_{\text {new }}}{(\cos \phi)_{\text {old }}}=\frac{1 / \sqrt{5}}{1 / \sqrt{10}}=\sqrt{\frac{10}{5}}=\sqrt{2}$ The ratio of new to old power factor is $\sqrt{2}$.
CG PET -2016
Alternating Current
155192
If $\mathrm{L}, \mathrm{C}, \mathrm{R}$ are respectively the inductance, capacitance and resistance, the quantities of dimensions same as of frequency are
1 $\frac{1}{\sqrt{\mathrm{LC}}}, \frac{\mathrm{R}}{\mathrm{L}}$ and $\frac{1}{\mathrm{RC}}$
2 $\sqrt{\mathrm{LC}}, \frac{\mathrm{L}}{\mathrm{R}}$ and $\mathrm{RC}$
3 $\sqrt{\frac{\mathrm{L}}{\mathrm{C}}}, \mathrm{LR}$ and $\frac{\mathrm{C}}{\mathrm{R}}$
4 $\sqrt{\frac{\mathrm{C}}{\mathrm{L}}}, \frac{1}{\mathrm{LR}}$ and $\frac{\mathrm{R}}{\mathrm{C}}$
Explanation:
A Dimension of $\frac{1}{\sqrt{\mathrm{LC}}}$ $=\frac{1}{\left[\mathrm{ML}^{2} \mathrm{~T}^{-2} \mathrm{~A}^{-2}\right]^{1 / 2}\left[\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^{4} \mathrm{~A}^{2}\right]^{1 / 2}}$ $=\frac{1}{[\mathrm{~T}]}=\left[\mathrm{T}^{-1}\right]$ Dimension of $\frac{\mathrm{R}}{\mathrm{L}}=\frac{\left[\mathrm{ML}^{2} \mathrm{~T}^{-3} \mathrm{~A}^{-2}\right]}{\left[\mathrm{ML}^{2} \mathrm{~T}^{-2} \mathrm{~A}^{-2}\right]}=\left[\mathrm{T}^{-1}\right]$ Dimension of $\frac{1}{\mathrm{RC}}=\frac{1}{\left[\mathrm{ML}^{2} \mathrm{~T}^{-3} \mathrm{~A}^{-2}\right]\left[\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^{4} \mathrm{~A}^{2}\right]}$ $=\left[\mathrm{T}^{-1}\right]$ Dimension of frequency $=\left[\mathrm{T}^{-1}\right]$ Hence, $\frac{1}{\sqrt{\mathrm{LC}}}, \frac{\mathrm{R}}{\mathrm{L}}$ and $\frac{1}{\mathrm{RC}}$ have the same dimension as frequency.
CG PET- 2011
Alternating Current
155193
$L-C-R$ circuit $V_{L}=V_{C}=V_{R}=10 V$ if $C$ is short circuited then voltage across $\mathrm{L}$ will be
1 $10 / \sqrt{2} \mathrm{~V}$
2 $20 \sqrt{2} \mathrm{~V}$
3 $10 \sqrt{2} \mathrm{~V}$
4 $10 \mathrm{~V}$
Explanation:
A Given, $V_{L}=V_{C}=V_{R}=10 \mathrm{~V}$ Then source voltage is given by - $\mathrm{V}=\sqrt{\mathrm{V}_{\mathrm{R}}^{2}+\left(\mathrm{V}_{\mathrm{L}}-\mathrm{V}_{\mathrm{C}}\right)^{2}}$ $\mathrm{~V}=\sqrt{(10)^{2}+(10-10)^{2}}$ $\mathrm{~V}=10 \mathrm{~V}$ Now if $\mathrm{C}$ is short circuited then voltage across capacitor is zero $\mathrm{V}_{\mathrm{C}}=0$ $\because \quad \mathrm{V}_{\mathrm{L}}, \mathrm{V}_{\mathrm{C}}$ and $\mathrm{V}_{\mathrm{R}}$ are equal $\text { And } \quad I=\frac{V}{\sqrt{R^{2}+X_{L}^{2}}}=\frac{10}{\sqrt{R^{2}+R^{2}}}=\frac{10}{\sqrt{2} R}$ Voltage across inductor is - $\mathrm{V}_{\mathrm{L}}=\mathrm{IX}_{\mathrm{L}}=\frac{10}{\sqrt{2} \mathrm{R}} \cdot \mathrm{R} \quad\left\{\because \mathrm{X}_{\mathrm{L}}=\mathrm{R}\right\}$ $\mathrm{V}_{\mathrm{L}}=\frac{10}{\sqrt{2}} \mathrm{~V}$
CG PET- 2007
Alternating Current
155194
The impedance of an $\mathrm{AC}$ circuit containing a capacitive reactance of $5 \Omega$ and inductive reactance of $8 \Omega$ will be
1 $1.6 \Omega$
2 $40 \Omega$
3 $3 \Omega$
4 $13 \Omega$
Explanation:
C Given that, Capacitive reactance $\mathrm{X}_{\mathrm{C}}=5 \Omega$, Inductive reactance $\mathrm{X}_{\mathrm{L}}=8 \Omega$, Impedance of $\mathrm{LC}$ circuits $Z=\sqrt{\left(X_{L}-X_{C}\right)^{2}}=\sqrt{(8-5)^{2}}=\sqrt{9}=3$ $Z=3 \Omega$
155190
In a circuit $L, C$ and $R$ are connected in series with an alternating voltage source of frequency f. The current leads the voltage by $45^{\circ}$. The value of $C$ is
C Given that, Phase difference between current and voltage of an series L-C-R circuit is $=45^{\circ}$ We know that, $\tan \theta=\frac{X_{L}-X_{C}}{R}$ $\tan 45^{\circ}=\frac{X_{L}-X_{C}}{R}$ $1=\frac{X_{L}-X_{C}}{R}$ $R=X_{L}-X_{C}$ $=2 \pi f L-\frac{1}{2 \pi f C}$ $\frac{1}{2 \pi f C}=2 \pi f L-R$ $C=\frac{1}{2 \pi f(2 \pi f L-R)}$
Manipal UGET-2010
Alternating Current
155191
In series $L R$ circuit $X_{L}=3 R$, now a capacitor with $X_{C}=R$ is added in series. The ratio of new to old power factor is
1 1
2 2
3 $\frac{1}{\sqrt{2}}$
4 $\sqrt{2}$
Explanation:
D Given, $X_{L}=3 R, X_{C}=R$ Condition 1 - power factor for $\mathrm{R}-\mathrm{C}$ circuit is - $(\cos \phi)_{\text {old }}=\frac{\mathrm{R}}{\mathrm{Z}}$ Impedance, $\mathrm{Z}=\sqrt{\mathrm{X}_{\mathrm{L}}^{2}+\mathrm{R}^{2}}$ $Z=\sqrt{(3 R)^{2}+R^{2}}$ $Z=\sqrt{10} R$ $(\cos \phi)_{\text {old }}=\frac{R}{\sqrt{10} R}=\frac{1}{\sqrt{10}}$ Condition 2 - If $X_{C}=R$ is added then Impedance, $Z_{\text {new }}=\sqrt{\left(X_{L}-X_{C}\right)^{2}+R^{2}}$ $=\sqrt{R^{2}+(3 R-R)^{2}}=\sqrt{5} R$ New power factor, $(\cos \phi)_{\text {new }}=\frac{\mathrm{R}}{\sqrt{5} \mathrm{R}}=\frac{1}{\sqrt{5}}$ $\frac{(\cos \phi)_{\text {new }}}{(\cos \phi)_{\text {old }}}=\frac{1 / \sqrt{5}}{1 / \sqrt{10}}=\sqrt{\frac{10}{5}}=\sqrt{2}$ The ratio of new to old power factor is $\sqrt{2}$.
CG PET -2016
Alternating Current
155192
If $\mathrm{L}, \mathrm{C}, \mathrm{R}$ are respectively the inductance, capacitance and resistance, the quantities of dimensions same as of frequency are
1 $\frac{1}{\sqrt{\mathrm{LC}}}, \frac{\mathrm{R}}{\mathrm{L}}$ and $\frac{1}{\mathrm{RC}}$
2 $\sqrt{\mathrm{LC}}, \frac{\mathrm{L}}{\mathrm{R}}$ and $\mathrm{RC}$
3 $\sqrt{\frac{\mathrm{L}}{\mathrm{C}}}, \mathrm{LR}$ and $\frac{\mathrm{C}}{\mathrm{R}}$
4 $\sqrt{\frac{\mathrm{C}}{\mathrm{L}}}, \frac{1}{\mathrm{LR}}$ and $\frac{\mathrm{R}}{\mathrm{C}}$
Explanation:
A Dimension of $\frac{1}{\sqrt{\mathrm{LC}}}$ $=\frac{1}{\left[\mathrm{ML}^{2} \mathrm{~T}^{-2} \mathrm{~A}^{-2}\right]^{1 / 2}\left[\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^{4} \mathrm{~A}^{2}\right]^{1 / 2}}$ $=\frac{1}{[\mathrm{~T}]}=\left[\mathrm{T}^{-1}\right]$ Dimension of $\frac{\mathrm{R}}{\mathrm{L}}=\frac{\left[\mathrm{ML}^{2} \mathrm{~T}^{-3} \mathrm{~A}^{-2}\right]}{\left[\mathrm{ML}^{2} \mathrm{~T}^{-2} \mathrm{~A}^{-2}\right]}=\left[\mathrm{T}^{-1}\right]$ Dimension of $\frac{1}{\mathrm{RC}}=\frac{1}{\left[\mathrm{ML}^{2} \mathrm{~T}^{-3} \mathrm{~A}^{-2}\right]\left[\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^{4} \mathrm{~A}^{2}\right]}$ $=\left[\mathrm{T}^{-1}\right]$ Dimension of frequency $=\left[\mathrm{T}^{-1}\right]$ Hence, $\frac{1}{\sqrt{\mathrm{LC}}}, \frac{\mathrm{R}}{\mathrm{L}}$ and $\frac{1}{\mathrm{RC}}$ have the same dimension as frequency.
CG PET- 2011
Alternating Current
155193
$L-C-R$ circuit $V_{L}=V_{C}=V_{R}=10 V$ if $C$ is short circuited then voltage across $\mathrm{L}$ will be
1 $10 / \sqrt{2} \mathrm{~V}$
2 $20 \sqrt{2} \mathrm{~V}$
3 $10 \sqrt{2} \mathrm{~V}$
4 $10 \mathrm{~V}$
Explanation:
A Given, $V_{L}=V_{C}=V_{R}=10 \mathrm{~V}$ Then source voltage is given by - $\mathrm{V}=\sqrt{\mathrm{V}_{\mathrm{R}}^{2}+\left(\mathrm{V}_{\mathrm{L}}-\mathrm{V}_{\mathrm{C}}\right)^{2}}$ $\mathrm{~V}=\sqrt{(10)^{2}+(10-10)^{2}}$ $\mathrm{~V}=10 \mathrm{~V}$ Now if $\mathrm{C}$ is short circuited then voltage across capacitor is zero $\mathrm{V}_{\mathrm{C}}=0$ $\because \quad \mathrm{V}_{\mathrm{L}}, \mathrm{V}_{\mathrm{C}}$ and $\mathrm{V}_{\mathrm{R}}$ are equal $\text { And } \quad I=\frac{V}{\sqrt{R^{2}+X_{L}^{2}}}=\frac{10}{\sqrt{R^{2}+R^{2}}}=\frac{10}{\sqrt{2} R}$ Voltage across inductor is - $\mathrm{V}_{\mathrm{L}}=\mathrm{IX}_{\mathrm{L}}=\frac{10}{\sqrt{2} \mathrm{R}} \cdot \mathrm{R} \quad\left\{\because \mathrm{X}_{\mathrm{L}}=\mathrm{R}\right\}$ $\mathrm{V}_{\mathrm{L}}=\frac{10}{\sqrt{2}} \mathrm{~V}$
CG PET- 2007
Alternating Current
155194
The impedance of an $\mathrm{AC}$ circuit containing a capacitive reactance of $5 \Omega$ and inductive reactance of $8 \Omega$ will be
1 $1.6 \Omega$
2 $40 \Omega$
3 $3 \Omega$
4 $13 \Omega$
Explanation:
C Given that, Capacitive reactance $\mathrm{X}_{\mathrm{C}}=5 \Omega$, Inductive reactance $\mathrm{X}_{\mathrm{L}}=8 \Omega$, Impedance of $\mathrm{LC}$ circuits $Z=\sqrt{\left(X_{L}-X_{C}\right)^{2}}=\sqrt{(8-5)^{2}}=\sqrt{9}=3$ $Z=3 \Omega$
155190
In a circuit $L, C$ and $R$ are connected in series with an alternating voltage source of frequency f. The current leads the voltage by $45^{\circ}$. The value of $C$ is
C Given that, Phase difference between current and voltage of an series L-C-R circuit is $=45^{\circ}$ We know that, $\tan \theta=\frac{X_{L}-X_{C}}{R}$ $\tan 45^{\circ}=\frac{X_{L}-X_{C}}{R}$ $1=\frac{X_{L}-X_{C}}{R}$ $R=X_{L}-X_{C}$ $=2 \pi f L-\frac{1}{2 \pi f C}$ $\frac{1}{2 \pi f C}=2 \pi f L-R$ $C=\frac{1}{2 \pi f(2 \pi f L-R)}$
Manipal UGET-2010
Alternating Current
155191
In series $L R$ circuit $X_{L}=3 R$, now a capacitor with $X_{C}=R$ is added in series. The ratio of new to old power factor is
1 1
2 2
3 $\frac{1}{\sqrt{2}}$
4 $\sqrt{2}$
Explanation:
D Given, $X_{L}=3 R, X_{C}=R$ Condition 1 - power factor for $\mathrm{R}-\mathrm{C}$ circuit is - $(\cos \phi)_{\text {old }}=\frac{\mathrm{R}}{\mathrm{Z}}$ Impedance, $\mathrm{Z}=\sqrt{\mathrm{X}_{\mathrm{L}}^{2}+\mathrm{R}^{2}}$ $Z=\sqrt{(3 R)^{2}+R^{2}}$ $Z=\sqrt{10} R$ $(\cos \phi)_{\text {old }}=\frac{R}{\sqrt{10} R}=\frac{1}{\sqrt{10}}$ Condition 2 - If $X_{C}=R$ is added then Impedance, $Z_{\text {new }}=\sqrt{\left(X_{L}-X_{C}\right)^{2}+R^{2}}$ $=\sqrt{R^{2}+(3 R-R)^{2}}=\sqrt{5} R$ New power factor, $(\cos \phi)_{\text {new }}=\frac{\mathrm{R}}{\sqrt{5} \mathrm{R}}=\frac{1}{\sqrt{5}}$ $\frac{(\cos \phi)_{\text {new }}}{(\cos \phi)_{\text {old }}}=\frac{1 / \sqrt{5}}{1 / \sqrt{10}}=\sqrt{\frac{10}{5}}=\sqrt{2}$ The ratio of new to old power factor is $\sqrt{2}$.
CG PET -2016
Alternating Current
155192
If $\mathrm{L}, \mathrm{C}, \mathrm{R}$ are respectively the inductance, capacitance and resistance, the quantities of dimensions same as of frequency are
1 $\frac{1}{\sqrt{\mathrm{LC}}}, \frac{\mathrm{R}}{\mathrm{L}}$ and $\frac{1}{\mathrm{RC}}$
2 $\sqrt{\mathrm{LC}}, \frac{\mathrm{L}}{\mathrm{R}}$ and $\mathrm{RC}$
3 $\sqrt{\frac{\mathrm{L}}{\mathrm{C}}}, \mathrm{LR}$ and $\frac{\mathrm{C}}{\mathrm{R}}$
4 $\sqrt{\frac{\mathrm{C}}{\mathrm{L}}}, \frac{1}{\mathrm{LR}}$ and $\frac{\mathrm{R}}{\mathrm{C}}$
Explanation:
A Dimension of $\frac{1}{\sqrt{\mathrm{LC}}}$ $=\frac{1}{\left[\mathrm{ML}^{2} \mathrm{~T}^{-2} \mathrm{~A}^{-2}\right]^{1 / 2}\left[\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^{4} \mathrm{~A}^{2}\right]^{1 / 2}}$ $=\frac{1}{[\mathrm{~T}]}=\left[\mathrm{T}^{-1}\right]$ Dimension of $\frac{\mathrm{R}}{\mathrm{L}}=\frac{\left[\mathrm{ML}^{2} \mathrm{~T}^{-3} \mathrm{~A}^{-2}\right]}{\left[\mathrm{ML}^{2} \mathrm{~T}^{-2} \mathrm{~A}^{-2}\right]}=\left[\mathrm{T}^{-1}\right]$ Dimension of $\frac{1}{\mathrm{RC}}=\frac{1}{\left[\mathrm{ML}^{2} \mathrm{~T}^{-3} \mathrm{~A}^{-2}\right]\left[\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^{4} \mathrm{~A}^{2}\right]}$ $=\left[\mathrm{T}^{-1}\right]$ Dimension of frequency $=\left[\mathrm{T}^{-1}\right]$ Hence, $\frac{1}{\sqrt{\mathrm{LC}}}, \frac{\mathrm{R}}{\mathrm{L}}$ and $\frac{1}{\mathrm{RC}}$ have the same dimension as frequency.
CG PET- 2011
Alternating Current
155193
$L-C-R$ circuit $V_{L}=V_{C}=V_{R}=10 V$ if $C$ is short circuited then voltage across $\mathrm{L}$ will be
1 $10 / \sqrt{2} \mathrm{~V}$
2 $20 \sqrt{2} \mathrm{~V}$
3 $10 \sqrt{2} \mathrm{~V}$
4 $10 \mathrm{~V}$
Explanation:
A Given, $V_{L}=V_{C}=V_{R}=10 \mathrm{~V}$ Then source voltage is given by - $\mathrm{V}=\sqrt{\mathrm{V}_{\mathrm{R}}^{2}+\left(\mathrm{V}_{\mathrm{L}}-\mathrm{V}_{\mathrm{C}}\right)^{2}}$ $\mathrm{~V}=\sqrt{(10)^{2}+(10-10)^{2}}$ $\mathrm{~V}=10 \mathrm{~V}$ Now if $\mathrm{C}$ is short circuited then voltage across capacitor is zero $\mathrm{V}_{\mathrm{C}}=0$ $\because \quad \mathrm{V}_{\mathrm{L}}, \mathrm{V}_{\mathrm{C}}$ and $\mathrm{V}_{\mathrm{R}}$ are equal $\text { And } \quad I=\frac{V}{\sqrt{R^{2}+X_{L}^{2}}}=\frac{10}{\sqrt{R^{2}+R^{2}}}=\frac{10}{\sqrt{2} R}$ Voltage across inductor is - $\mathrm{V}_{\mathrm{L}}=\mathrm{IX}_{\mathrm{L}}=\frac{10}{\sqrt{2} \mathrm{R}} \cdot \mathrm{R} \quad\left\{\because \mathrm{X}_{\mathrm{L}}=\mathrm{R}\right\}$ $\mathrm{V}_{\mathrm{L}}=\frac{10}{\sqrt{2}} \mathrm{~V}$
CG PET- 2007
Alternating Current
155194
The impedance of an $\mathrm{AC}$ circuit containing a capacitive reactance of $5 \Omega$ and inductive reactance of $8 \Omega$ will be
1 $1.6 \Omega$
2 $40 \Omega$
3 $3 \Omega$
4 $13 \Omega$
Explanation:
C Given that, Capacitive reactance $\mathrm{X}_{\mathrm{C}}=5 \Omega$, Inductive reactance $\mathrm{X}_{\mathrm{L}}=8 \Omega$, Impedance of $\mathrm{LC}$ circuits $Z=\sqrt{\left(X_{L}-X_{C}\right)^{2}}=\sqrt{(8-5)^{2}}=\sqrt{9}=3$ $Z=3 \Omega$