359307
Three plates of common surface area \(A\) are connected as shown. The effective capacitance will be
1 \(\frac{{2{\varepsilon _o}A}}{d}\)
2 \(\frac{3}{2}\frac{{{\varepsilon _o}A}}{d}\)
3 \(\frac{{3{\varepsilon _o}A}}{d}\)
4 \(\frac{{{\varepsilon _o}A}}{d}\)
Explanation:
The given circuit is equivalent to parallel combination of two identical capacitors, each having capacitance \(C = \frac{{{\varepsilon _o}A}}{d}\) Hence \({C_{eq}} = 2C = \frac{{2{\varepsilon _o}A}}{d}\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359308
Four identical capacitors are connected as shown in daigram. When a battery of 6\(V\) is connected between \(A\) and \(B\), the charge stored is found to be \(1.5\mu C\) . The value of \(C\) is:
1 \(15\mu F\)
2 \(2.5\mu F\)
3 \(0.1\mu F\)
4 \(1.5\mu F\)
Explanation:
The capacitance across \(A\) and \(B\) \( = \frac{C}{2} + C + C = \frac{5}{2}C\) As \(Q = {C_{AB}}V,\) \(1.5\mu C = \frac{5}{2}C \times 6\) \( \Rightarrow C = 0.1\mu F\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359309
The combined capacity of the parallel combination of two capacitors is four times their combined capacity when connected in series. This means that
1 Their capacities are equal and \(2\mu F\) each
2 Their capacities are \(1\mu F\;{\rm{and}}\;2\mu F\)
3 Their capacities are \(0.5\mu F\;{\rm{and}}\;1\mu F\)
4 Their capacities are infinite
Explanation:
Let \(C = 2\mu F\) each \({C_{Parallel}} = 4\mu F\) \({C_{Series}} = 1\mu F\) \( \Rightarrow \) Option (1) is correct.
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359310
Five capacitors each of capacity \(C\) are connected as shown in figure. If their resultant capacity is \(2\mu F\) , then the capacity of each condenser is
1 \(5\mu F\)
2 \(2\mu F\)
3 \(2.5\mu F\)
4 \(10\mu F\)
Explanation:
Given, resultant capacity \({C_{eq}} = 2\mu F\) The equivelent capacity of the circuit is given by \(\frac{1}{{{C_{eq}}}} = \frac{1}{C} + \frac{1}{C} + \frac{1}{C} + \frac{1}{C} + \frac{1}{C}\) ( \(\therefore \) all capacitors are in series) \( \Rightarrow \frac{1}{{{C_{eq}}}} = \frac{5}{C}\) \( \Rightarrow \frac{1}{{2\mu F}} = \frac{5}{C}\) \( \Rightarrow C = 10\mu F\)
359307
Three plates of common surface area \(A\) are connected as shown. The effective capacitance will be
1 \(\frac{{2{\varepsilon _o}A}}{d}\)
2 \(\frac{3}{2}\frac{{{\varepsilon _o}A}}{d}\)
3 \(\frac{{3{\varepsilon _o}A}}{d}\)
4 \(\frac{{{\varepsilon _o}A}}{d}\)
Explanation:
The given circuit is equivalent to parallel combination of two identical capacitors, each having capacitance \(C = \frac{{{\varepsilon _o}A}}{d}\) Hence \({C_{eq}} = 2C = \frac{{2{\varepsilon _o}A}}{d}\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359308
Four identical capacitors are connected as shown in daigram. When a battery of 6\(V\) is connected between \(A\) and \(B\), the charge stored is found to be \(1.5\mu C\) . The value of \(C\) is:
1 \(15\mu F\)
2 \(2.5\mu F\)
3 \(0.1\mu F\)
4 \(1.5\mu F\)
Explanation:
The capacitance across \(A\) and \(B\) \( = \frac{C}{2} + C + C = \frac{5}{2}C\) As \(Q = {C_{AB}}V,\) \(1.5\mu C = \frac{5}{2}C \times 6\) \( \Rightarrow C = 0.1\mu F\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359309
The combined capacity of the parallel combination of two capacitors is four times their combined capacity when connected in series. This means that
1 Their capacities are equal and \(2\mu F\) each
2 Their capacities are \(1\mu F\;{\rm{and}}\;2\mu F\)
3 Their capacities are \(0.5\mu F\;{\rm{and}}\;1\mu F\)
4 Their capacities are infinite
Explanation:
Let \(C = 2\mu F\) each \({C_{Parallel}} = 4\mu F\) \({C_{Series}} = 1\mu F\) \( \Rightarrow \) Option (1) is correct.
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359310
Five capacitors each of capacity \(C\) are connected as shown in figure. If their resultant capacity is \(2\mu F\) , then the capacity of each condenser is
1 \(5\mu F\)
2 \(2\mu F\)
3 \(2.5\mu F\)
4 \(10\mu F\)
Explanation:
Given, resultant capacity \({C_{eq}} = 2\mu F\) The equivelent capacity of the circuit is given by \(\frac{1}{{{C_{eq}}}} = \frac{1}{C} + \frac{1}{C} + \frac{1}{C} + \frac{1}{C} + \frac{1}{C}\) ( \(\therefore \) all capacitors are in series) \( \Rightarrow \frac{1}{{{C_{eq}}}} = \frac{5}{C}\) \( \Rightarrow \frac{1}{{2\mu F}} = \frac{5}{C}\) \( \Rightarrow C = 10\mu F\)
359307
Three plates of common surface area \(A\) are connected as shown. The effective capacitance will be
1 \(\frac{{2{\varepsilon _o}A}}{d}\)
2 \(\frac{3}{2}\frac{{{\varepsilon _o}A}}{d}\)
3 \(\frac{{3{\varepsilon _o}A}}{d}\)
4 \(\frac{{{\varepsilon _o}A}}{d}\)
Explanation:
The given circuit is equivalent to parallel combination of two identical capacitors, each having capacitance \(C = \frac{{{\varepsilon _o}A}}{d}\) Hence \({C_{eq}} = 2C = \frac{{2{\varepsilon _o}A}}{d}\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359308
Four identical capacitors are connected as shown in daigram. When a battery of 6\(V\) is connected between \(A\) and \(B\), the charge stored is found to be \(1.5\mu C\) . The value of \(C\) is:
1 \(15\mu F\)
2 \(2.5\mu F\)
3 \(0.1\mu F\)
4 \(1.5\mu F\)
Explanation:
The capacitance across \(A\) and \(B\) \( = \frac{C}{2} + C + C = \frac{5}{2}C\) As \(Q = {C_{AB}}V,\) \(1.5\mu C = \frac{5}{2}C \times 6\) \( \Rightarrow C = 0.1\mu F\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359309
The combined capacity of the parallel combination of two capacitors is four times their combined capacity when connected in series. This means that
1 Their capacities are equal and \(2\mu F\) each
2 Their capacities are \(1\mu F\;{\rm{and}}\;2\mu F\)
3 Their capacities are \(0.5\mu F\;{\rm{and}}\;1\mu F\)
4 Their capacities are infinite
Explanation:
Let \(C = 2\mu F\) each \({C_{Parallel}} = 4\mu F\) \({C_{Series}} = 1\mu F\) \( \Rightarrow \) Option (1) is correct.
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359310
Five capacitors each of capacity \(C\) are connected as shown in figure. If their resultant capacity is \(2\mu F\) , then the capacity of each condenser is
1 \(5\mu F\)
2 \(2\mu F\)
3 \(2.5\mu F\)
4 \(10\mu F\)
Explanation:
Given, resultant capacity \({C_{eq}} = 2\mu F\) The equivelent capacity of the circuit is given by \(\frac{1}{{{C_{eq}}}} = \frac{1}{C} + \frac{1}{C} + \frac{1}{C} + \frac{1}{C} + \frac{1}{C}\) ( \(\therefore \) all capacitors are in series) \( \Rightarrow \frac{1}{{{C_{eq}}}} = \frac{5}{C}\) \( \Rightarrow \frac{1}{{2\mu F}} = \frac{5}{C}\) \( \Rightarrow C = 10\mu F\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359307
Three plates of common surface area \(A\) are connected as shown. The effective capacitance will be
1 \(\frac{{2{\varepsilon _o}A}}{d}\)
2 \(\frac{3}{2}\frac{{{\varepsilon _o}A}}{d}\)
3 \(\frac{{3{\varepsilon _o}A}}{d}\)
4 \(\frac{{{\varepsilon _o}A}}{d}\)
Explanation:
The given circuit is equivalent to parallel combination of two identical capacitors, each having capacitance \(C = \frac{{{\varepsilon _o}A}}{d}\) Hence \({C_{eq}} = 2C = \frac{{2{\varepsilon _o}A}}{d}\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359308
Four identical capacitors are connected as shown in daigram. When a battery of 6\(V\) is connected between \(A\) and \(B\), the charge stored is found to be \(1.5\mu C\) . The value of \(C\) is:
1 \(15\mu F\)
2 \(2.5\mu F\)
3 \(0.1\mu F\)
4 \(1.5\mu F\)
Explanation:
The capacitance across \(A\) and \(B\) \( = \frac{C}{2} + C + C = \frac{5}{2}C\) As \(Q = {C_{AB}}V,\) \(1.5\mu C = \frac{5}{2}C \times 6\) \( \Rightarrow C = 0.1\mu F\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359309
The combined capacity of the parallel combination of two capacitors is four times their combined capacity when connected in series. This means that
1 Their capacities are equal and \(2\mu F\) each
2 Their capacities are \(1\mu F\;{\rm{and}}\;2\mu F\)
3 Their capacities are \(0.5\mu F\;{\rm{and}}\;1\mu F\)
4 Their capacities are infinite
Explanation:
Let \(C = 2\mu F\) each \({C_{Parallel}} = 4\mu F\) \({C_{Series}} = 1\mu F\) \( \Rightarrow \) Option (1) is correct.
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359310
Five capacitors each of capacity \(C\) are connected as shown in figure. If their resultant capacity is \(2\mu F\) , then the capacity of each condenser is
1 \(5\mu F\)
2 \(2\mu F\)
3 \(2.5\mu F\)
4 \(10\mu F\)
Explanation:
Given, resultant capacity \({C_{eq}} = 2\mu F\) The equivelent capacity of the circuit is given by \(\frac{1}{{{C_{eq}}}} = \frac{1}{C} + \frac{1}{C} + \frac{1}{C} + \frac{1}{C} + \frac{1}{C}\) ( \(\therefore \) all capacitors are in series) \( \Rightarrow \frac{1}{{{C_{eq}}}} = \frac{5}{C}\) \( \Rightarrow \frac{1}{{2\mu F}} = \frac{5}{C}\) \( \Rightarrow C = 10\mu F\)
