165844
In the given figure, the charge on $3 \mu \mathrm{F}$ capacitor is
1 $10 \mu \mathrm{C}$
2 $15 \mu \mathrm{C}$
3 $30 \mu \mathrm{C}$
4 $5 \mu \mathrm{C}$
Explanation:
: According to Figure, all the capacitors are connected in series, Hence, net capacitance is, $\mathrm{C}_{\mathrm{net}}=\frac{1}{\mathrm{C}_{1}}+\frac{1}{\mathrm{C}_{2}}+\frac{1}{\mathrm{C}_{3}}$ $\frac{1}{\mathrm{C}}=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}$ $\frac{1}{\mathrm{C}}=\frac{3+2+1}{6}=\frac{6}{6}$ $\mathrm{C}_{\text {net }}=1 \mu \mathrm{F}$ Being series each capacitor is same charge, Hence, $\mathrm{Q} =\mathrm{C}_{\text {net }} \cdot \mathrm{V} \quad(\mathrm{V}=10 \mathrm{~V})$ $=1 \times 10$ $\mathrm{Q} =10 \mu$
[BITSAT-2019]
Capacitance
165845
In the given circuit the potential drop across the capacitor must be
1 $\mathrm{V}$
2 $\frac{\mathrm{V}}{2}$
3 $\frac{\mathrm{V}}{3}$
4 $\frac{2 \mathrm{~V}}{3}$
Explanation:
: Given, In steady state, no current will pass through the capacitor, In the lower loop, $2 \mathrm{~V}-2 \mathrm{iR}-\mathrm{iR}-\mathrm{V}=0$ $\mathrm{i}=\mathrm{V} / 3 \mathrm{R}$ For the upper loop, $V-V_{C}-i R-V=0$ $V_{C}=i R$ $\left|V_{C}\right|=\frac{V}{3 R} \times R$ $\left|V_{C}\right|=\frac{V}{3}$
[BITSAT-2016]
Capacitance
165846
In the figure below, what is the potential difference between the point $A$ and $B$ and between $B$ and $C$ respectively in steady state
: Given, At steady, current through capacitors is zero. Hence effect of resistors in the circuit can be neglected. When capacitors are connected in series with a battery of potential V. Potential difference across each capacitor is $\mathrm{V}_{1}=\left[\frac{\mathrm{C}_{2}}{\mathrm{C}_{1}+\mathrm{C}_{2}}\right] \mathrm{V}$ $\mathrm{V}_{2}=\left[\frac{\mathrm{C}_{1}}{\mathrm{C}_{1}+\mathrm{C}_{2}}\right] \mathrm{V}$ Now, $\mathrm{V}_{1}=\mathrm{V}_{\mathrm{AB}}=\left[\frac{2}{2+6}\right] \times 100=25 \mathrm{~V}$ $\mathrm{~V}_{2}=\mathrm{V}_{\mathrm{BC}}=\left[\frac{6}{2+6}\right] \times 100=75 \mathrm{~V}$ So, correct answer is option (c).
165844
In the given figure, the charge on $3 \mu \mathrm{F}$ capacitor is
1 $10 \mu \mathrm{C}$
2 $15 \mu \mathrm{C}$
3 $30 \mu \mathrm{C}$
4 $5 \mu \mathrm{C}$
Explanation:
: According to Figure, all the capacitors are connected in series, Hence, net capacitance is, $\mathrm{C}_{\mathrm{net}}=\frac{1}{\mathrm{C}_{1}}+\frac{1}{\mathrm{C}_{2}}+\frac{1}{\mathrm{C}_{3}}$ $\frac{1}{\mathrm{C}}=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}$ $\frac{1}{\mathrm{C}}=\frac{3+2+1}{6}=\frac{6}{6}$ $\mathrm{C}_{\text {net }}=1 \mu \mathrm{F}$ Being series each capacitor is same charge, Hence, $\mathrm{Q} =\mathrm{C}_{\text {net }} \cdot \mathrm{V} \quad(\mathrm{V}=10 \mathrm{~V})$ $=1 \times 10$ $\mathrm{Q} =10 \mu$
[BITSAT-2019]
Capacitance
165845
In the given circuit the potential drop across the capacitor must be
1 $\mathrm{V}$
2 $\frac{\mathrm{V}}{2}$
3 $\frac{\mathrm{V}}{3}$
4 $\frac{2 \mathrm{~V}}{3}$
Explanation:
: Given, In steady state, no current will pass through the capacitor, In the lower loop, $2 \mathrm{~V}-2 \mathrm{iR}-\mathrm{iR}-\mathrm{V}=0$ $\mathrm{i}=\mathrm{V} / 3 \mathrm{R}$ For the upper loop, $V-V_{C}-i R-V=0$ $V_{C}=i R$ $\left|V_{C}\right|=\frac{V}{3 R} \times R$ $\left|V_{C}\right|=\frac{V}{3}$
[BITSAT-2016]
Capacitance
165846
In the figure below, what is the potential difference between the point $A$ and $B$ and between $B$ and $C$ respectively in steady state
: Given, At steady, current through capacitors is zero. Hence effect of resistors in the circuit can be neglected. When capacitors are connected in series with a battery of potential V. Potential difference across each capacitor is $\mathrm{V}_{1}=\left[\frac{\mathrm{C}_{2}}{\mathrm{C}_{1}+\mathrm{C}_{2}}\right] \mathrm{V}$ $\mathrm{V}_{2}=\left[\frac{\mathrm{C}_{1}}{\mathrm{C}_{1}+\mathrm{C}_{2}}\right] \mathrm{V}$ Now, $\mathrm{V}_{1}=\mathrm{V}_{\mathrm{AB}}=\left[\frac{2}{2+6}\right] \times 100=25 \mathrm{~V}$ $\mathrm{~V}_{2}=\mathrm{V}_{\mathrm{BC}}=\left[\frac{6}{2+6}\right] \times 100=75 \mathrm{~V}$ So, correct answer is option (c).
165844
In the given figure, the charge on $3 \mu \mathrm{F}$ capacitor is
1 $10 \mu \mathrm{C}$
2 $15 \mu \mathrm{C}$
3 $30 \mu \mathrm{C}$
4 $5 \mu \mathrm{C}$
Explanation:
: According to Figure, all the capacitors are connected in series, Hence, net capacitance is, $\mathrm{C}_{\mathrm{net}}=\frac{1}{\mathrm{C}_{1}}+\frac{1}{\mathrm{C}_{2}}+\frac{1}{\mathrm{C}_{3}}$ $\frac{1}{\mathrm{C}}=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}$ $\frac{1}{\mathrm{C}}=\frac{3+2+1}{6}=\frac{6}{6}$ $\mathrm{C}_{\text {net }}=1 \mu \mathrm{F}$ Being series each capacitor is same charge, Hence, $\mathrm{Q} =\mathrm{C}_{\text {net }} \cdot \mathrm{V} \quad(\mathrm{V}=10 \mathrm{~V})$ $=1 \times 10$ $\mathrm{Q} =10 \mu$
[BITSAT-2019]
Capacitance
165845
In the given circuit the potential drop across the capacitor must be
1 $\mathrm{V}$
2 $\frac{\mathrm{V}}{2}$
3 $\frac{\mathrm{V}}{3}$
4 $\frac{2 \mathrm{~V}}{3}$
Explanation:
: Given, In steady state, no current will pass through the capacitor, In the lower loop, $2 \mathrm{~V}-2 \mathrm{iR}-\mathrm{iR}-\mathrm{V}=0$ $\mathrm{i}=\mathrm{V} / 3 \mathrm{R}$ For the upper loop, $V-V_{C}-i R-V=0$ $V_{C}=i R$ $\left|V_{C}\right|=\frac{V}{3 R} \times R$ $\left|V_{C}\right|=\frac{V}{3}$
[BITSAT-2016]
Capacitance
165846
In the figure below, what is the potential difference between the point $A$ and $B$ and between $B$ and $C$ respectively in steady state
: Given, At steady, current through capacitors is zero. Hence effect of resistors in the circuit can be neglected. When capacitors are connected in series with a battery of potential V. Potential difference across each capacitor is $\mathrm{V}_{1}=\left[\frac{\mathrm{C}_{2}}{\mathrm{C}_{1}+\mathrm{C}_{2}}\right] \mathrm{V}$ $\mathrm{V}_{2}=\left[\frac{\mathrm{C}_{1}}{\mathrm{C}_{1}+\mathrm{C}_{2}}\right] \mathrm{V}$ Now, $\mathrm{V}_{1}=\mathrm{V}_{\mathrm{AB}}=\left[\frac{2}{2+6}\right] \times 100=25 \mathrm{~V}$ $\mathrm{~V}_{2}=\mathrm{V}_{\mathrm{BC}}=\left[\frac{6}{2+6}\right] \times 100=75 \mathrm{~V}$ So, correct answer is option (c).
165844
In the given figure, the charge on $3 \mu \mathrm{F}$ capacitor is
1 $10 \mu \mathrm{C}$
2 $15 \mu \mathrm{C}$
3 $30 \mu \mathrm{C}$
4 $5 \mu \mathrm{C}$
Explanation:
: According to Figure, all the capacitors are connected in series, Hence, net capacitance is, $\mathrm{C}_{\mathrm{net}}=\frac{1}{\mathrm{C}_{1}}+\frac{1}{\mathrm{C}_{2}}+\frac{1}{\mathrm{C}_{3}}$ $\frac{1}{\mathrm{C}}=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}$ $\frac{1}{\mathrm{C}}=\frac{3+2+1}{6}=\frac{6}{6}$ $\mathrm{C}_{\text {net }}=1 \mu \mathrm{F}$ Being series each capacitor is same charge, Hence, $\mathrm{Q} =\mathrm{C}_{\text {net }} \cdot \mathrm{V} \quad(\mathrm{V}=10 \mathrm{~V})$ $=1 \times 10$ $\mathrm{Q} =10 \mu$
[BITSAT-2019]
Capacitance
165845
In the given circuit the potential drop across the capacitor must be
1 $\mathrm{V}$
2 $\frac{\mathrm{V}}{2}$
3 $\frac{\mathrm{V}}{3}$
4 $\frac{2 \mathrm{~V}}{3}$
Explanation:
: Given, In steady state, no current will pass through the capacitor, In the lower loop, $2 \mathrm{~V}-2 \mathrm{iR}-\mathrm{iR}-\mathrm{V}=0$ $\mathrm{i}=\mathrm{V} / 3 \mathrm{R}$ For the upper loop, $V-V_{C}-i R-V=0$ $V_{C}=i R$ $\left|V_{C}\right|=\frac{V}{3 R} \times R$ $\left|V_{C}\right|=\frac{V}{3}$
[BITSAT-2016]
Capacitance
165846
In the figure below, what is the potential difference between the point $A$ and $B$ and between $B$ and $C$ respectively in steady state
: Given, At steady, current through capacitors is zero. Hence effect of resistors in the circuit can be neglected. When capacitors are connected in series with a battery of potential V. Potential difference across each capacitor is $\mathrm{V}_{1}=\left[\frac{\mathrm{C}_{2}}{\mathrm{C}_{1}+\mathrm{C}_{2}}\right] \mathrm{V}$ $\mathrm{V}_{2}=\left[\frac{\mathrm{C}_{1}}{\mathrm{C}_{1}+\mathrm{C}_{2}}\right] \mathrm{V}$ Now, $\mathrm{V}_{1}=\mathrm{V}_{\mathrm{AB}}=\left[\frac{2}{2+6}\right] \times 100=25 \mathrm{~V}$ $\mathrm{~V}_{2}=\mathrm{V}_{\mathrm{BC}}=\left[\frac{6}{2+6}\right] \times 100=75 \mathrm{~V}$ So, correct answer is option (c).
