NEET Test Series from KOTA - 10 Papers In MS WORD
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Current Electricity
153101
A $20 \mu \mathrm{F}$ capacitor is connected to $45 \mathrm{~V}$ battery through a circuit whose resistance is $2000 \Omega$. What is the final charge on the capacitor?
1 $9 \times 10^{-4} \mathrm{C}$
2 $9.154 \times 10^{-4} \mathrm{C}$
3 $9.8 \times 10^{-4} \mathrm{C}$
4 None of these
Explanation:
A Given, $\mathrm{C}=20 \mu \mathrm{F}, \mathrm{V}=45 \mathrm{~V}$ We have that, the potential difference across the capacitor $=45 \mathrm{~V}$ Hence, the final charge on the capacitor is $\mathrm{q}=\mathrm{CV}$ $q=20 \times 10^{-6} \times 45$ $q=900 \times 10^{-6}$ $q=9 \times 10^{-4} \mathrm{C}$
UPSEE - 2008
Current Electricity
153102
The charge $q$ in the circuit shown here varies with time $t$ is as :
1 a
2 b
3 c
4 d
Explanation:
A The given is a resistor capacitor circuit or RC network. It is one of the simplest analogue electronic filters. The equation governing the charging of capacitor C through resistance $\mathrm{R}$ is $\mathrm{q}=\mathrm{q}_{0}\left(1-\mathrm{e}^{\mathrm{t} / \mathrm{R}}\right)$. It shows that the charge on the capacitor grows exponentially, reaching its steady value $\mathrm{q}_{0}=\mathrm{CE}$ asymptotically at $\mathrm{t} \rightarrow \infty\left(\mathrm{e}^{-\infty}=0\right)$. A graph between charge and time is
UPSEE - 2005
Current Electricity
153104
In the given figure the steady state current in the circuit is-
1 zero
2 $0.6 \mathrm{~A}$
3 $0.9 \mathrm{~A}$
4 $1.5 \mathrm{~A}$
Explanation:
D Given that, circuit the resistors of $2 \Omega$ and $3 \Omega$ are connected in parallel $\frac{1}{\mathrm{R}^{\prime}}=\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}$ $=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}$ $\mathrm{R}^{\prime}=\frac{6}{5}=1.2 \Omega$ $\mathrm{R}^{\prime \prime}=1.2+2.8 \Omega$ $=4 \Omega$ $\mathrm{i}=\frac{\mathrm{V}}{\mathrm{R}} \cdot=\frac{6}{4}=1.5 \mathrm{~A}$
BCECE-2007
Current Electricity
153105
When the key $K$ is pressed at time $t=0$, which of the following is true about the current $I$ in the resistor $\mathrm{AB}$ ?
1 $2 \mathrm{~mA}$ all the time
2 oscillates between $1 \mathrm{~mA}$ and $2 \mathrm{~mA}$
3 $1 \mathrm{~mA}$ all the time
4 At $\mathrm{t}=0, \mathrm{I}=2 \mathrm{~mA}$ and reduces to $1 \mathrm{~mA}$ finally
Explanation:
D At time $\mathrm{t}=0$ When capacitor is charging current $\mathrm{I}=\frac{2}{1000}=2 \mathrm{~mA}$ New resistance of circuit $(R)=1+1=2 \mathrm{k} \Omega$ When capacitor is full charged no current will pass through it hence current through the circuit $I =\frac{2}{2000}$ $=1 \mathrm{~mA}$
153101
A $20 \mu \mathrm{F}$ capacitor is connected to $45 \mathrm{~V}$ battery through a circuit whose resistance is $2000 \Omega$. What is the final charge on the capacitor?
1 $9 \times 10^{-4} \mathrm{C}$
2 $9.154 \times 10^{-4} \mathrm{C}$
3 $9.8 \times 10^{-4} \mathrm{C}$
4 None of these
Explanation:
A Given, $\mathrm{C}=20 \mu \mathrm{F}, \mathrm{V}=45 \mathrm{~V}$ We have that, the potential difference across the capacitor $=45 \mathrm{~V}$ Hence, the final charge on the capacitor is $\mathrm{q}=\mathrm{CV}$ $q=20 \times 10^{-6} \times 45$ $q=900 \times 10^{-6}$ $q=9 \times 10^{-4} \mathrm{C}$
UPSEE - 2008
Current Electricity
153102
The charge $q$ in the circuit shown here varies with time $t$ is as :
1 a
2 b
3 c
4 d
Explanation:
A The given is a resistor capacitor circuit or RC network. It is one of the simplest analogue electronic filters. The equation governing the charging of capacitor C through resistance $\mathrm{R}$ is $\mathrm{q}=\mathrm{q}_{0}\left(1-\mathrm{e}^{\mathrm{t} / \mathrm{R}}\right)$. It shows that the charge on the capacitor grows exponentially, reaching its steady value $\mathrm{q}_{0}=\mathrm{CE}$ asymptotically at $\mathrm{t} \rightarrow \infty\left(\mathrm{e}^{-\infty}=0\right)$. A graph between charge and time is
UPSEE - 2005
Current Electricity
153104
In the given figure the steady state current in the circuit is-
1 zero
2 $0.6 \mathrm{~A}$
3 $0.9 \mathrm{~A}$
4 $1.5 \mathrm{~A}$
Explanation:
D Given that, circuit the resistors of $2 \Omega$ and $3 \Omega$ are connected in parallel $\frac{1}{\mathrm{R}^{\prime}}=\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}$ $=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}$ $\mathrm{R}^{\prime}=\frac{6}{5}=1.2 \Omega$ $\mathrm{R}^{\prime \prime}=1.2+2.8 \Omega$ $=4 \Omega$ $\mathrm{i}=\frac{\mathrm{V}}{\mathrm{R}} \cdot=\frac{6}{4}=1.5 \mathrm{~A}$
BCECE-2007
Current Electricity
153105
When the key $K$ is pressed at time $t=0$, which of the following is true about the current $I$ in the resistor $\mathrm{AB}$ ?
1 $2 \mathrm{~mA}$ all the time
2 oscillates between $1 \mathrm{~mA}$ and $2 \mathrm{~mA}$
3 $1 \mathrm{~mA}$ all the time
4 At $\mathrm{t}=0, \mathrm{I}=2 \mathrm{~mA}$ and reduces to $1 \mathrm{~mA}$ finally
Explanation:
D At time $\mathrm{t}=0$ When capacitor is charging current $\mathrm{I}=\frac{2}{1000}=2 \mathrm{~mA}$ New resistance of circuit $(R)=1+1=2 \mathrm{k} \Omega$ When capacitor is full charged no current will pass through it hence current through the circuit $I =\frac{2}{2000}$ $=1 \mathrm{~mA}$
153101
A $20 \mu \mathrm{F}$ capacitor is connected to $45 \mathrm{~V}$ battery through a circuit whose resistance is $2000 \Omega$. What is the final charge on the capacitor?
1 $9 \times 10^{-4} \mathrm{C}$
2 $9.154 \times 10^{-4} \mathrm{C}$
3 $9.8 \times 10^{-4} \mathrm{C}$
4 None of these
Explanation:
A Given, $\mathrm{C}=20 \mu \mathrm{F}, \mathrm{V}=45 \mathrm{~V}$ We have that, the potential difference across the capacitor $=45 \mathrm{~V}$ Hence, the final charge on the capacitor is $\mathrm{q}=\mathrm{CV}$ $q=20 \times 10^{-6} \times 45$ $q=900 \times 10^{-6}$ $q=9 \times 10^{-4} \mathrm{C}$
UPSEE - 2008
Current Electricity
153102
The charge $q$ in the circuit shown here varies with time $t$ is as :
1 a
2 b
3 c
4 d
Explanation:
A The given is a resistor capacitor circuit or RC network. It is one of the simplest analogue electronic filters. The equation governing the charging of capacitor C through resistance $\mathrm{R}$ is $\mathrm{q}=\mathrm{q}_{0}\left(1-\mathrm{e}^{\mathrm{t} / \mathrm{R}}\right)$. It shows that the charge on the capacitor grows exponentially, reaching its steady value $\mathrm{q}_{0}=\mathrm{CE}$ asymptotically at $\mathrm{t} \rightarrow \infty\left(\mathrm{e}^{-\infty}=0\right)$. A graph between charge and time is
UPSEE - 2005
Current Electricity
153104
In the given figure the steady state current in the circuit is-
1 zero
2 $0.6 \mathrm{~A}$
3 $0.9 \mathrm{~A}$
4 $1.5 \mathrm{~A}$
Explanation:
D Given that, circuit the resistors of $2 \Omega$ and $3 \Omega$ are connected in parallel $\frac{1}{\mathrm{R}^{\prime}}=\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}$ $=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}$ $\mathrm{R}^{\prime}=\frac{6}{5}=1.2 \Omega$ $\mathrm{R}^{\prime \prime}=1.2+2.8 \Omega$ $=4 \Omega$ $\mathrm{i}=\frac{\mathrm{V}}{\mathrm{R}} \cdot=\frac{6}{4}=1.5 \mathrm{~A}$
BCECE-2007
Current Electricity
153105
When the key $K$ is pressed at time $t=0$, which of the following is true about the current $I$ in the resistor $\mathrm{AB}$ ?
1 $2 \mathrm{~mA}$ all the time
2 oscillates between $1 \mathrm{~mA}$ and $2 \mathrm{~mA}$
3 $1 \mathrm{~mA}$ all the time
4 At $\mathrm{t}=0, \mathrm{I}=2 \mathrm{~mA}$ and reduces to $1 \mathrm{~mA}$ finally
Explanation:
D At time $\mathrm{t}=0$ When capacitor is charging current $\mathrm{I}=\frac{2}{1000}=2 \mathrm{~mA}$ New resistance of circuit $(R)=1+1=2 \mathrm{k} \Omega$ When capacitor is full charged no current will pass through it hence current through the circuit $I =\frac{2}{2000}$ $=1 \mathrm{~mA}$
153101
A $20 \mu \mathrm{F}$ capacitor is connected to $45 \mathrm{~V}$ battery through a circuit whose resistance is $2000 \Omega$. What is the final charge on the capacitor?
1 $9 \times 10^{-4} \mathrm{C}$
2 $9.154 \times 10^{-4} \mathrm{C}$
3 $9.8 \times 10^{-4} \mathrm{C}$
4 None of these
Explanation:
A Given, $\mathrm{C}=20 \mu \mathrm{F}, \mathrm{V}=45 \mathrm{~V}$ We have that, the potential difference across the capacitor $=45 \mathrm{~V}$ Hence, the final charge on the capacitor is $\mathrm{q}=\mathrm{CV}$ $q=20 \times 10^{-6} \times 45$ $q=900 \times 10^{-6}$ $q=9 \times 10^{-4} \mathrm{C}$
UPSEE - 2008
Current Electricity
153102
The charge $q$ in the circuit shown here varies with time $t$ is as :
1 a
2 b
3 c
4 d
Explanation:
A The given is a resistor capacitor circuit or RC network. It is one of the simplest analogue electronic filters. The equation governing the charging of capacitor C through resistance $\mathrm{R}$ is $\mathrm{q}=\mathrm{q}_{0}\left(1-\mathrm{e}^{\mathrm{t} / \mathrm{R}}\right)$. It shows that the charge on the capacitor grows exponentially, reaching its steady value $\mathrm{q}_{0}=\mathrm{CE}$ asymptotically at $\mathrm{t} \rightarrow \infty\left(\mathrm{e}^{-\infty}=0\right)$. A graph between charge and time is
UPSEE - 2005
Current Electricity
153104
In the given figure the steady state current in the circuit is-
1 zero
2 $0.6 \mathrm{~A}$
3 $0.9 \mathrm{~A}$
4 $1.5 \mathrm{~A}$
Explanation:
D Given that, circuit the resistors of $2 \Omega$ and $3 \Omega$ are connected in parallel $\frac{1}{\mathrm{R}^{\prime}}=\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}$ $=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}$ $\mathrm{R}^{\prime}=\frac{6}{5}=1.2 \Omega$ $\mathrm{R}^{\prime \prime}=1.2+2.8 \Omega$ $=4 \Omega$ $\mathrm{i}=\frac{\mathrm{V}}{\mathrm{R}} \cdot=\frac{6}{4}=1.5 \mathrm{~A}$
BCECE-2007
Current Electricity
153105
When the key $K$ is pressed at time $t=0$, which of the following is true about the current $I$ in the resistor $\mathrm{AB}$ ?
1 $2 \mathrm{~mA}$ all the time
2 oscillates between $1 \mathrm{~mA}$ and $2 \mathrm{~mA}$
3 $1 \mathrm{~mA}$ all the time
4 At $\mathrm{t}=0, \mathrm{I}=2 \mathrm{~mA}$ and reduces to $1 \mathrm{~mA}$ finally
Explanation:
D At time $\mathrm{t}=0$ When capacitor is charging current $\mathrm{I}=\frac{2}{1000}=2 \mathrm{~mA}$ New resistance of circuit $(R)=1+1=2 \mathrm{k} \Omega$ When capacitor is full charged no current will pass through it hence current through the circuit $I =\frac{2}{2000}$ $=1 \mathrm{~mA}$
