359294
If three capacitors each of capacity 1 \(mF\) are connected in such a way that the resultant capacity is 1.5 \(mF\), then
1 All the three are connected in series
2 All the three are connected in parallel
3 Two of them are in parallel and connected in series to the third
4 Two of them are in series and then connected in parallel to the third
Explanation:
Conceptual Question
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359295
The equivalent capacitance of the combination shown in the figure is :
1 \(2C\)
2 \(C/2\)
3 \(3C/2\)
4 \(3C\)
Explanation:
The middle capacitor is short circuted The circuit can be redrawn as \({C_{AB}} = 2C\)
NEET - 2021
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359296
A parallel plate capacitor is formed by stacking in equally spaced plates connected alternately. If capacitance between two adjacent plates is \(C\) then the resultant capacitance is
1 \(nC\)
2 \((n + 1)C\)
3 \((n - 1)C\)
4 \(C\)
Explanation:
\({C_{eq}} = \left( {n - 1} \right)C\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359297
Five identical capacitors, when connected in series, produce equivalent capacitance of \(6\mu F\). Now these capacitors are rearranged so as to produce the maximum value of equivalent capacitance. This maximum value will be :
1 \(36\mu F\)
2 \(30\mu F\)
3 \(6\mu F\)
4 \(150\mu F\)
Explanation:
Let \(C\) be the capacitance of each capacitor Then \(6 = \frac{C}{5}\) \( \Rightarrow C = 30\mu F\) If these are connected in parallel, then equivalent capacitance will be maximum. \(\therefore C' = 30 \times 5 = 150\mu F\)
359294
If three capacitors each of capacity 1 \(mF\) are connected in such a way that the resultant capacity is 1.5 \(mF\), then
1 All the three are connected in series
2 All the three are connected in parallel
3 Two of them are in parallel and connected in series to the third
4 Two of them are in series and then connected in parallel to the third
Explanation:
Conceptual Question
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359295
The equivalent capacitance of the combination shown in the figure is :
1 \(2C\)
2 \(C/2\)
3 \(3C/2\)
4 \(3C\)
Explanation:
The middle capacitor is short circuted The circuit can be redrawn as \({C_{AB}} = 2C\)
NEET - 2021
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359296
A parallel plate capacitor is formed by stacking in equally spaced plates connected alternately. If capacitance between two adjacent plates is \(C\) then the resultant capacitance is
1 \(nC\)
2 \((n + 1)C\)
3 \((n - 1)C\)
4 \(C\)
Explanation:
\({C_{eq}} = \left( {n - 1} \right)C\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359297
Five identical capacitors, when connected in series, produce equivalent capacitance of \(6\mu F\). Now these capacitors are rearranged so as to produce the maximum value of equivalent capacitance. This maximum value will be :
1 \(36\mu F\)
2 \(30\mu F\)
3 \(6\mu F\)
4 \(150\mu F\)
Explanation:
Let \(C\) be the capacitance of each capacitor Then \(6 = \frac{C}{5}\) \( \Rightarrow C = 30\mu F\) If these are connected in parallel, then equivalent capacitance will be maximum. \(\therefore C' = 30 \times 5 = 150\mu F\)
359294
If three capacitors each of capacity 1 \(mF\) are connected in such a way that the resultant capacity is 1.5 \(mF\), then
1 All the three are connected in series
2 All the three are connected in parallel
3 Two of them are in parallel and connected in series to the third
4 Two of them are in series and then connected in parallel to the third
Explanation:
Conceptual Question
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359295
The equivalent capacitance of the combination shown in the figure is :
1 \(2C\)
2 \(C/2\)
3 \(3C/2\)
4 \(3C\)
Explanation:
The middle capacitor is short circuted The circuit can be redrawn as \({C_{AB}} = 2C\)
NEET - 2021
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359296
A parallel plate capacitor is formed by stacking in equally spaced plates connected alternately. If capacitance between two adjacent plates is \(C\) then the resultant capacitance is
1 \(nC\)
2 \((n + 1)C\)
3 \((n - 1)C\)
4 \(C\)
Explanation:
\({C_{eq}} = \left( {n - 1} \right)C\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359297
Five identical capacitors, when connected in series, produce equivalent capacitance of \(6\mu F\). Now these capacitors are rearranged so as to produce the maximum value of equivalent capacitance. This maximum value will be :
1 \(36\mu F\)
2 \(30\mu F\)
3 \(6\mu F\)
4 \(150\mu F\)
Explanation:
Let \(C\) be the capacitance of each capacitor Then \(6 = \frac{C}{5}\) \( \Rightarrow C = 30\mu F\) If these are connected in parallel, then equivalent capacitance will be maximum. \(\therefore C' = 30 \times 5 = 150\mu F\)
359294
If three capacitors each of capacity 1 \(mF\) are connected in such a way that the resultant capacity is 1.5 \(mF\), then
1 All the three are connected in series
2 All the three are connected in parallel
3 Two of them are in parallel and connected in series to the third
4 Two of them are in series and then connected in parallel to the third
Explanation:
Conceptual Question
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359295
The equivalent capacitance of the combination shown in the figure is :
1 \(2C\)
2 \(C/2\)
3 \(3C/2\)
4 \(3C\)
Explanation:
The middle capacitor is short circuted The circuit can be redrawn as \({C_{AB}} = 2C\)
NEET - 2021
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359296
A parallel plate capacitor is formed by stacking in equally spaced plates connected alternately. If capacitance between two adjacent plates is \(C\) then the resultant capacitance is
1 \(nC\)
2 \((n + 1)C\)
3 \((n - 1)C\)
4 \(C\)
Explanation:
\({C_{eq}} = \left( {n - 1} \right)C\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359297
Five identical capacitors, when connected in series, produce equivalent capacitance of \(6\mu F\). Now these capacitors are rearranged so as to produce the maximum value of equivalent capacitance. This maximum value will be :
1 \(36\mu F\)
2 \(30\mu F\)
3 \(6\mu F\)
4 \(150\mu F\)
Explanation:
Let \(C\) be the capacitance of each capacitor Then \(6 = \frac{C}{5}\) \( \Rightarrow C = 30\mu F\) If these are connected in parallel, then equivalent capacitance will be maximum. \(\therefore C' = 30 \times 5 = 150\mu F\)
