153096
Given, $\mathrm{R}_{1}=1 \Omega \mathrm{C}_{1}=2 \mu \mathrm{F}$ $\mathrm{R}_{2}=2 \Omega \mathrm{C}_{2}=4 \mu \mathrm{F}$ The time constant (in $\mu$ s) for the circuits I, II, III are respectively
1 $18,8 / 9,4$
2 $18,4,8 / 9$
3 $4,8 / 9,18$
4 $8 / 9,18,4$
Explanation:
D Given, $\mathrm{R}_{1}=1 \Omega, \mathrm{R}_{2}=2 \Omega, \mathrm{C}_{1}=2 \mu \mathrm{F}, \mathrm{C}_{2}=4 \mu \mathrm{F}$ From figure (I) - Net capacitance $(C)=\frac{C_{1} C_{2}}{C_{1}+C_{2}}=\frac{2 \times 4}{2+4}=\frac{4}{3} \mu \mathrm{F}$ Net resistance $(\mathrm{R})=\frac{\mathrm{R}_{1} \mathrm{R}_{2}}{\mathrm{R}_{1}+\mathrm{R}_{2}}=\frac{1 \times 2}{1+2}=\frac{2}{3} \Omega$ $\because$ Time constants $(\tau)=\mathrm{CR}$ $=\frac{4}{3} \times \frac{2}{3}=\frac{8}{9} \mu \mathrm{s}$ From figure (II) - $\mathrm{C}=\mathrm{C}_{1}+\mathrm{C}_{2}=2+4=6 \mu \mathrm{F}$ $\mathrm{R}=\mathrm{R}_{1}+\mathrm{R}_{2}=1+2=3 \Omega$ Time constants $(\tau)=\mathrm{CR}$ From figure (III) - $=6 \times 3=18 \mu \mathrm{s}$ $\mathrm{C}=\mathrm{C}_{1}+\mathrm{C}_{2}=2+4=6 \mu \mathrm{F}$ $\mathrm{R}=\frac{\mathrm{R}_{1} \mathrm{R}_{2}}{\mathrm{R}_{1}+\mathrm{R}_{2}}=\frac{1 \times 2}{1+2}=\frac{2}{3} \Omega$ Time constants $(\tau)=6 \times \frac{2}{3}=4 \mu \mathrm{s}$
Manipal UGET-2014
Current Electricity
153097
The time constant of $L-R$ circuited is
1 $\mathrm{L} / \mathrm{R}$
2 $\mathrm{R} / \mathrm{L}$
3 $1 / \mathrm{RL}$
4 RL
Explanation:
A Voltage across inductor, $\mathrm{V}_{\mathrm{L}}(\mathrm{t})=\mathrm{V}_{0} \mathrm{e}^{\frac{-\mathrm{Rt}}{\mathrm{L}}}$ Where, $\mathrm{V}_{\mathrm{L}}=$ Voltage across the inductor $\mathrm{V}_{0}=$ Maximum achievable voltage $\mathrm{t}=$ Time elapsed after switching on the circuit $\mathrm{R}=$ Resistance of the circuit $\mathrm{L}=$ Inductance of the inductor Use the definition of time constant, $\text { At } \mathrm{t}=\tau$ $\mathrm{V}_{0} \mathrm{e}^{\frac{-\mathrm{R} \tau}{\mathrm{L}}}=\frac{\mathrm{V}_{0}}{\mathrm{e}}$ $\frac{-\mathrm{R} \tau}{\mathrm{L}}=-1$ $\tau=\frac{\mathrm{L}}{\mathrm{R}}$
CG PET- 2012
Current Electricity
153099
Two identical capacitors each of capacitance $C$ are charged to the same potential $V$ and are connected in two circuits (i) and (ii) at $t=0$ as shown. The charged on the capacitor at $t=C R$ are
C From figure (I), The p-n junction diode is forward biased and represent a very low resistance, the capacitor, therefore discharges itself through resistor $\mathrm{R}$ according to relation. $\text { and }\mathrm{q}=\mathrm{q}_{0} \mathrm{e}^{-\mathrm{t} C \mathrm{R}}$ $\mathrm{q}_{0}=\mathrm{CV} \text { at } \mathrm{t}=\mathrm{CR}$ $\therefore \mathrm{q}=\mathrm{q}_{0} \mathrm{e}^{-1}=\frac{\mathrm{CV}}{\mathrm{e}}$ From figure (II), In $\mathrm{p}-\mathrm{n}$ junction diode is reverse biased, the capacitor, therefore holds the charge intact, $\therefore \quad \mathrm{q}=\mathrm{q}_{0}=\mathrm{CV}$
UPSEE - 2012
Current Electricity
153100
Taking the internal resistance of the battery as negligible, the steady state current in the $2 \Omega$ resistor shown in the figure will be
1 $1.8 \mathrm{~A}$
2 $2.9 \mathrm{~A}$
3 $0.9 \mathrm{~A}$
4 $2.8 \mathrm{~A}$
Explanation:
C In the steady state, capacitor offers infinite resistance to DC, so no current flows through $6 \Omega$ resistor, which is thus ineffective. The equivalent circuit is shown below in which $2 \Omega$ and $3 \Omega$ are in parallel. $\frac{1}{\mathrm{R}_{\mathrm{eq}}} =\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}$ $=\frac{1}{2}+\frac{1}{3}=\frac{3+2}{6}$ $\mathrm{R}_{\mathrm{eq}}=\frac{6}{5}=1.2 \Omega$ $1.2 \Omega$ resistance is in series with $2.8 \Omega$ $\therefore \quad \mathrm{R}_{\mathrm{eq}}^{\prime}=1.2+2.8=4 \Omega$ Current drawn from the battery, $\mathrm{I}=\frac{\mathrm{E}}{\mathrm{R}}=\frac{6}{4}=1.5 \mathrm{~A}$ Potential difference between points $\mathrm{A}$ and $\mathrm{B}$ $\mathrm{V}=\mathrm{IR}^{\prime}=1.5 \times 1.2=1.8 \mathrm{~V}$ $\therefore$ Current through $2 \Omega$ resistor $=\frac{\mathrm{V}}{2}=\frac{1.8}{2}=0.9 \mathrm{~A}$
153096
Given, $\mathrm{R}_{1}=1 \Omega \mathrm{C}_{1}=2 \mu \mathrm{F}$ $\mathrm{R}_{2}=2 \Omega \mathrm{C}_{2}=4 \mu \mathrm{F}$ The time constant (in $\mu$ s) for the circuits I, II, III are respectively
1 $18,8 / 9,4$
2 $18,4,8 / 9$
3 $4,8 / 9,18$
4 $8 / 9,18,4$
Explanation:
D Given, $\mathrm{R}_{1}=1 \Omega, \mathrm{R}_{2}=2 \Omega, \mathrm{C}_{1}=2 \mu \mathrm{F}, \mathrm{C}_{2}=4 \mu \mathrm{F}$ From figure (I) - Net capacitance $(C)=\frac{C_{1} C_{2}}{C_{1}+C_{2}}=\frac{2 \times 4}{2+4}=\frac{4}{3} \mu \mathrm{F}$ Net resistance $(\mathrm{R})=\frac{\mathrm{R}_{1} \mathrm{R}_{2}}{\mathrm{R}_{1}+\mathrm{R}_{2}}=\frac{1 \times 2}{1+2}=\frac{2}{3} \Omega$ $\because$ Time constants $(\tau)=\mathrm{CR}$ $=\frac{4}{3} \times \frac{2}{3}=\frac{8}{9} \mu \mathrm{s}$ From figure (II) - $\mathrm{C}=\mathrm{C}_{1}+\mathrm{C}_{2}=2+4=6 \mu \mathrm{F}$ $\mathrm{R}=\mathrm{R}_{1}+\mathrm{R}_{2}=1+2=3 \Omega$ Time constants $(\tau)=\mathrm{CR}$ From figure (III) - $=6 \times 3=18 \mu \mathrm{s}$ $\mathrm{C}=\mathrm{C}_{1}+\mathrm{C}_{2}=2+4=6 \mu \mathrm{F}$ $\mathrm{R}=\frac{\mathrm{R}_{1} \mathrm{R}_{2}}{\mathrm{R}_{1}+\mathrm{R}_{2}}=\frac{1 \times 2}{1+2}=\frac{2}{3} \Omega$ Time constants $(\tau)=6 \times \frac{2}{3}=4 \mu \mathrm{s}$
Manipal UGET-2014
Current Electricity
153097
The time constant of $L-R$ circuited is
1 $\mathrm{L} / \mathrm{R}$
2 $\mathrm{R} / \mathrm{L}$
3 $1 / \mathrm{RL}$
4 RL
Explanation:
A Voltage across inductor, $\mathrm{V}_{\mathrm{L}}(\mathrm{t})=\mathrm{V}_{0} \mathrm{e}^{\frac{-\mathrm{Rt}}{\mathrm{L}}}$ Where, $\mathrm{V}_{\mathrm{L}}=$ Voltage across the inductor $\mathrm{V}_{0}=$ Maximum achievable voltage $\mathrm{t}=$ Time elapsed after switching on the circuit $\mathrm{R}=$ Resistance of the circuit $\mathrm{L}=$ Inductance of the inductor Use the definition of time constant, $\text { At } \mathrm{t}=\tau$ $\mathrm{V}_{0} \mathrm{e}^{\frac{-\mathrm{R} \tau}{\mathrm{L}}}=\frac{\mathrm{V}_{0}}{\mathrm{e}}$ $\frac{-\mathrm{R} \tau}{\mathrm{L}}=-1$ $\tau=\frac{\mathrm{L}}{\mathrm{R}}$
CG PET- 2012
Current Electricity
153099
Two identical capacitors each of capacitance $C$ are charged to the same potential $V$ and are connected in two circuits (i) and (ii) at $t=0$ as shown. The charged on the capacitor at $t=C R$ are
C From figure (I), The p-n junction diode is forward biased and represent a very low resistance, the capacitor, therefore discharges itself through resistor $\mathrm{R}$ according to relation. $\text { and }\mathrm{q}=\mathrm{q}_{0} \mathrm{e}^{-\mathrm{t} C \mathrm{R}}$ $\mathrm{q}_{0}=\mathrm{CV} \text { at } \mathrm{t}=\mathrm{CR}$ $\therefore \mathrm{q}=\mathrm{q}_{0} \mathrm{e}^{-1}=\frac{\mathrm{CV}}{\mathrm{e}}$ From figure (II), In $\mathrm{p}-\mathrm{n}$ junction diode is reverse biased, the capacitor, therefore holds the charge intact, $\therefore \quad \mathrm{q}=\mathrm{q}_{0}=\mathrm{CV}$
UPSEE - 2012
Current Electricity
153100
Taking the internal resistance of the battery as negligible, the steady state current in the $2 \Omega$ resistor shown in the figure will be
1 $1.8 \mathrm{~A}$
2 $2.9 \mathrm{~A}$
3 $0.9 \mathrm{~A}$
4 $2.8 \mathrm{~A}$
Explanation:
C In the steady state, capacitor offers infinite resistance to DC, so no current flows through $6 \Omega$ resistor, which is thus ineffective. The equivalent circuit is shown below in which $2 \Omega$ and $3 \Omega$ are in parallel. $\frac{1}{\mathrm{R}_{\mathrm{eq}}} =\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}$ $=\frac{1}{2}+\frac{1}{3}=\frac{3+2}{6}$ $\mathrm{R}_{\mathrm{eq}}=\frac{6}{5}=1.2 \Omega$ $1.2 \Omega$ resistance is in series with $2.8 \Omega$ $\therefore \quad \mathrm{R}_{\mathrm{eq}}^{\prime}=1.2+2.8=4 \Omega$ Current drawn from the battery, $\mathrm{I}=\frac{\mathrm{E}}{\mathrm{R}}=\frac{6}{4}=1.5 \mathrm{~A}$ Potential difference between points $\mathrm{A}$ and $\mathrm{B}$ $\mathrm{V}=\mathrm{IR}^{\prime}=1.5 \times 1.2=1.8 \mathrm{~V}$ $\therefore$ Current through $2 \Omega$ resistor $=\frac{\mathrm{V}}{2}=\frac{1.8}{2}=0.9 \mathrm{~A}$
153096
Given, $\mathrm{R}_{1}=1 \Omega \mathrm{C}_{1}=2 \mu \mathrm{F}$ $\mathrm{R}_{2}=2 \Omega \mathrm{C}_{2}=4 \mu \mathrm{F}$ The time constant (in $\mu$ s) for the circuits I, II, III are respectively
1 $18,8 / 9,4$
2 $18,4,8 / 9$
3 $4,8 / 9,18$
4 $8 / 9,18,4$
Explanation:
D Given, $\mathrm{R}_{1}=1 \Omega, \mathrm{R}_{2}=2 \Omega, \mathrm{C}_{1}=2 \mu \mathrm{F}, \mathrm{C}_{2}=4 \mu \mathrm{F}$ From figure (I) - Net capacitance $(C)=\frac{C_{1} C_{2}}{C_{1}+C_{2}}=\frac{2 \times 4}{2+4}=\frac{4}{3} \mu \mathrm{F}$ Net resistance $(\mathrm{R})=\frac{\mathrm{R}_{1} \mathrm{R}_{2}}{\mathrm{R}_{1}+\mathrm{R}_{2}}=\frac{1 \times 2}{1+2}=\frac{2}{3} \Omega$ $\because$ Time constants $(\tau)=\mathrm{CR}$ $=\frac{4}{3} \times \frac{2}{3}=\frac{8}{9} \mu \mathrm{s}$ From figure (II) - $\mathrm{C}=\mathrm{C}_{1}+\mathrm{C}_{2}=2+4=6 \mu \mathrm{F}$ $\mathrm{R}=\mathrm{R}_{1}+\mathrm{R}_{2}=1+2=3 \Omega$ Time constants $(\tau)=\mathrm{CR}$ From figure (III) - $=6 \times 3=18 \mu \mathrm{s}$ $\mathrm{C}=\mathrm{C}_{1}+\mathrm{C}_{2}=2+4=6 \mu \mathrm{F}$ $\mathrm{R}=\frac{\mathrm{R}_{1} \mathrm{R}_{2}}{\mathrm{R}_{1}+\mathrm{R}_{2}}=\frac{1 \times 2}{1+2}=\frac{2}{3} \Omega$ Time constants $(\tau)=6 \times \frac{2}{3}=4 \mu \mathrm{s}$
Manipal UGET-2014
Current Electricity
153097
The time constant of $L-R$ circuited is
1 $\mathrm{L} / \mathrm{R}$
2 $\mathrm{R} / \mathrm{L}$
3 $1 / \mathrm{RL}$
4 RL
Explanation:
A Voltage across inductor, $\mathrm{V}_{\mathrm{L}}(\mathrm{t})=\mathrm{V}_{0} \mathrm{e}^{\frac{-\mathrm{Rt}}{\mathrm{L}}}$ Where, $\mathrm{V}_{\mathrm{L}}=$ Voltage across the inductor $\mathrm{V}_{0}=$ Maximum achievable voltage $\mathrm{t}=$ Time elapsed after switching on the circuit $\mathrm{R}=$ Resistance of the circuit $\mathrm{L}=$ Inductance of the inductor Use the definition of time constant, $\text { At } \mathrm{t}=\tau$ $\mathrm{V}_{0} \mathrm{e}^{\frac{-\mathrm{R} \tau}{\mathrm{L}}}=\frac{\mathrm{V}_{0}}{\mathrm{e}}$ $\frac{-\mathrm{R} \tau}{\mathrm{L}}=-1$ $\tau=\frac{\mathrm{L}}{\mathrm{R}}$
CG PET- 2012
Current Electricity
153099
Two identical capacitors each of capacitance $C$ are charged to the same potential $V$ and are connected in two circuits (i) and (ii) at $t=0$ as shown. The charged on the capacitor at $t=C R$ are
C From figure (I), The p-n junction diode is forward biased and represent a very low resistance, the capacitor, therefore discharges itself through resistor $\mathrm{R}$ according to relation. $\text { and }\mathrm{q}=\mathrm{q}_{0} \mathrm{e}^{-\mathrm{t} C \mathrm{R}}$ $\mathrm{q}_{0}=\mathrm{CV} \text { at } \mathrm{t}=\mathrm{CR}$ $\therefore \mathrm{q}=\mathrm{q}_{0} \mathrm{e}^{-1}=\frac{\mathrm{CV}}{\mathrm{e}}$ From figure (II), In $\mathrm{p}-\mathrm{n}$ junction diode is reverse biased, the capacitor, therefore holds the charge intact, $\therefore \quad \mathrm{q}=\mathrm{q}_{0}=\mathrm{CV}$
UPSEE - 2012
Current Electricity
153100
Taking the internal resistance of the battery as negligible, the steady state current in the $2 \Omega$ resistor shown in the figure will be
1 $1.8 \mathrm{~A}$
2 $2.9 \mathrm{~A}$
3 $0.9 \mathrm{~A}$
4 $2.8 \mathrm{~A}$
Explanation:
C In the steady state, capacitor offers infinite resistance to DC, so no current flows through $6 \Omega$ resistor, which is thus ineffective. The equivalent circuit is shown below in which $2 \Omega$ and $3 \Omega$ are in parallel. $\frac{1}{\mathrm{R}_{\mathrm{eq}}} =\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}$ $=\frac{1}{2}+\frac{1}{3}=\frac{3+2}{6}$ $\mathrm{R}_{\mathrm{eq}}=\frac{6}{5}=1.2 \Omega$ $1.2 \Omega$ resistance is in series with $2.8 \Omega$ $\therefore \quad \mathrm{R}_{\mathrm{eq}}^{\prime}=1.2+2.8=4 \Omega$ Current drawn from the battery, $\mathrm{I}=\frac{\mathrm{E}}{\mathrm{R}}=\frac{6}{4}=1.5 \mathrm{~A}$ Potential difference between points $\mathrm{A}$ and $\mathrm{B}$ $\mathrm{V}=\mathrm{IR}^{\prime}=1.5 \times 1.2=1.8 \mathrm{~V}$ $\therefore$ Current through $2 \Omega$ resistor $=\frac{\mathrm{V}}{2}=\frac{1.8}{2}=0.9 \mathrm{~A}$
153096
Given, $\mathrm{R}_{1}=1 \Omega \mathrm{C}_{1}=2 \mu \mathrm{F}$ $\mathrm{R}_{2}=2 \Omega \mathrm{C}_{2}=4 \mu \mathrm{F}$ The time constant (in $\mu$ s) for the circuits I, II, III are respectively
1 $18,8 / 9,4$
2 $18,4,8 / 9$
3 $4,8 / 9,18$
4 $8 / 9,18,4$
Explanation:
D Given, $\mathrm{R}_{1}=1 \Omega, \mathrm{R}_{2}=2 \Omega, \mathrm{C}_{1}=2 \mu \mathrm{F}, \mathrm{C}_{2}=4 \mu \mathrm{F}$ From figure (I) - Net capacitance $(C)=\frac{C_{1} C_{2}}{C_{1}+C_{2}}=\frac{2 \times 4}{2+4}=\frac{4}{3} \mu \mathrm{F}$ Net resistance $(\mathrm{R})=\frac{\mathrm{R}_{1} \mathrm{R}_{2}}{\mathrm{R}_{1}+\mathrm{R}_{2}}=\frac{1 \times 2}{1+2}=\frac{2}{3} \Omega$ $\because$ Time constants $(\tau)=\mathrm{CR}$ $=\frac{4}{3} \times \frac{2}{3}=\frac{8}{9} \mu \mathrm{s}$ From figure (II) - $\mathrm{C}=\mathrm{C}_{1}+\mathrm{C}_{2}=2+4=6 \mu \mathrm{F}$ $\mathrm{R}=\mathrm{R}_{1}+\mathrm{R}_{2}=1+2=3 \Omega$ Time constants $(\tau)=\mathrm{CR}$ From figure (III) - $=6 \times 3=18 \mu \mathrm{s}$ $\mathrm{C}=\mathrm{C}_{1}+\mathrm{C}_{2}=2+4=6 \mu \mathrm{F}$ $\mathrm{R}=\frac{\mathrm{R}_{1} \mathrm{R}_{2}}{\mathrm{R}_{1}+\mathrm{R}_{2}}=\frac{1 \times 2}{1+2}=\frac{2}{3} \Omega$ Time constants $(\tau)=6 \times \frac{2}{3}=4 \mu \mathrm{s}$
Manipal UGET-2014
Current Electricity
153097
The time constant of $L-R$ circuited is
1 $\mathrm{L} / \mathrm{R}$
2 $\mathrm{R} / \mathrm{L}$
3 $1 / \mathrm{RL}$
4 RL
Explanation:
A Voltage across inductor, $\mathrm{V}_{\mathrm{L}}(\mathrm{t})=\mathrm{V}_{0} \mathrm{e}^{\frac{-\mathrm{Rt}}{\mathrm{L}}}$ Where, $\mathrm{V}_{\mathrm{L}}=$ Voltage across the inductor $\mathrm{V}_{0}=$ Maximum achievable voltage $\mathrm{t}=$ Time elapsed after switching on the circuit $\mathrm{R}=$ Resistance of the circuit $\mathrm{L}=$ Inductance of the inductor Use the definition of time constant, $\text { At } \mathrm{t}=\tau$ $\mathrm{V}_{0} \mathrm{e}^{\frac{-\mathrm{R} \tau}{\mathrm{L}}}=\frac{\mathrm{V}_{0}}{\mathrm{e}}$ $\frac{-\mathrm{R} \tau}{\mathrm{L}}=-1$ $\tau=\frac{\mathrm{L}}{\mathrm{R}}$
CG PET- 2012
Current Electricity
153099
Two identical capacitors each of capacitance $C$ are charged to the same potential $V$ and are connected in two circuits (i) and (ii) at $t=0$ as shown. The charged on the capacitor at $t=C R$ are
C From figure (I), The p-n junction diode is forward biased and represent a very low resistance, the capacitor, therefore discharges itself through resistor $\mathrm{R}$ according to relation. $\text { and }\mathrm{q}=\mathrm{q}_{0} \mathrm{e}^{-\mathrm{t} C \mathrm{R}}$ $\mathrm{q}_{0}=\mathrm{CV} \text { at } \mathrm{t}=\mathrm{CR}$ $\therefore \mathrm{q}=\mathrm{q}_{0} \mathrm{e}^{-1}=\frac{\mathrm{CV}}{\mathrm{e}}$ From figure (II), In $\mathrm{p}-\mathrm{n}$ junction diode is reverse biased, the capacitor, therefore holds the charge intact, $\therefore \quad \mathrm{q}=\mathrm{q}_{0}=\mathrm{CV}$
UPSEE - 2012
Current Electricity
153100
Taking the internal resistance of the battery as negligible, the steady state current in the $2 \Omega$ resistor shown in the figure will be
1 $1.8 \mathrm{~A}$
2 $2.9 \mathrm{~A}$
3 $0.9 \mathrm{~A}$
4 $2.8 \mathrm{~A}$
Explanation:
C In the steady state, capacitor offers infinite resistance to DC, so no current flows through $6 \Omega$ resistor, which is thus ineffective. The equivalent circuit is shown below in which $2 \Omega$ and $3 \Omega$ are in parallel. $\frac{1}{\mathrm{R}_{\mathrm{eq}}} =\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}$ $=\frac{1}{2}+\frac{1}{3}=\frac{3+2}{6}$ $\mathrm{R}_{\mathrm{eq}}=\frac{6}{5}=1.2 \Omega$ $1.2 \Omega$ resistance is in series with $2.8 \Omega$ $\therefore \quad \mathrm{R}_{\mathrm{eq}}^{\prime}=1.2+2.8=4 \Omega$ Current drawn from the battery, $\mathrm{I}=\frac{\mathrm{E}}{\mathrm{R}}=\frac{6}{4}=1.5 \mathrm{~A}$ Potential difference between points $\mathrm{A}$ and $\mathrm{B}$ $\mathrm{V}=\mathrm{IR}^{\prime}=1.5 \times 1.2=1.8 \mathrm{~V}$ $\therefore$ Current through $2 \Omega$ resistor $=\frac{\mathrm{V}}{2}=\frac{1.8}{2}=0.9 \mathrm{~A}$
