NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359286
In figure below, four parallel plates of equal area \(A\) and spacing \(d\) are arrange, then effective capacitance between \(a\) and \(b\) is
1 \(\dfrac{\varepsilon_{0} A}{d}\)
2 \(\dfrac{2 \varepsilon_{0} A}{d}\)
3 \(\dfrac{3 \varepsilon_{0} A}{d}\)
4 \(\dfrac{4 \varepsilon_{0} A}{d}\)
Explanation:
Suppose the pair of plates \(a\) is connected to positive terminal of the battery and the pair of plate \(b\) is connected to negative terminal of the battery. Thus, from figure, we can conclude that, we have three capacitors I, II and III.
Their positive plates are connected to each other and so are the negative plates. Therefore, these are joined in parallel,\({\text{ }}i.e\). \(C_{p}=C_{1}+C_{2}+C_{3}=3 \varepsilon_{0} A / d\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359287
When two identical capacitors are in series, they have \(4\,\mu F\) capacitance and when in parallel \(16\,\mu F\). What is the capacitance of each?
1 \(6\,\mu F\)
2 \(4\,\mu F\)
3 \(8\,\mu F\)
4 \(16\,\mu F\)
Explanation:
Let the arrangement be shown as and \({\therefore \quad \dfrac{1}{C_{\text {eq }}}=\dfrac{1}{C}+\dfrac{1}{C}}\) \({\therefore C_{\text {eq }}=C+C}\) \({C_{e q}=\dfrac{C^{2}}{2 C}}\) \({16 \mu {F}=2 {C}}\) \({\Rightarrow 4 \mu {F}=\dfrac{{C}^{2}}{2 C}}\) \({\therefore {C}=8 \mu {F}}\) \({\Rightarrow {C}=8 \mu {F}}\). So correct option is (3)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359288
All capacitors used in the diagram are identical and each is of capacitance \(C\). Then the effective capacitance between the points \(A\) and \(B\) is
1 \(3C\)
2 \(6C\)
3 \(1.5C\)
4 \(C\)
Explanation:
Let \(P\) be a point. The points having identical potentials are shown in the figure. The circuit can be redrawn as \({C_{AB}} = 1.5C.\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359289
The equivalent capacitance between points \({A}\) and \({B}\) of circuit shown in the figure is
1 \({9 \mu {F}}\)
2 \({1 \mu {F}}\)
3 \({4.5 \mu {F}}\)
4 \({6 \mu {F}}\)
Explanation:
Since, it is a parallel combination of all three capacitors. \({C_{eq}} = C + C + C = 3C = 3 \times 3\mu F,{C_{eq}} = 9\mu F.\) So correct option is (1)
359286
In figure below, four parallel plates of equal area \(A\) and spacing \(d\) are arrange, then effective capacitance between \(a\) and \(b\) is
1 \(\dfrac{\varepsilon_{0} A}{d}\)
2 \(\dfrac{2 \varepsilon_{0} A}{d}\)
3 \(\dfrac{3 \varepsilon_{0} A}{d}\)
4 \(\dfrac{4 \varepsilon_{0} A}{d}\)
Explanation:
Suppose the pair of plates \(a\) is connected to positive terminal of the battery and the pair of plate \(b\) is connected to negative terminal of the battery. Thus, from figure, we can conclude that, we have three capacitors I, II and III.
Their positive plates are connected to each other and so are the negative plates. Therefore, these are joined in parallel,\({\text{ }}i.e\). \(C_{p}=C_{1}+C_{2}+C_{3}=3 \varepsilon_{0} A / d\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359287
When two identical capacitors are in series, they have \(4\,\mu F\) capacitance and when in parallel \(16\,\mu F\). What is the capacitance of each?
1 \(6\,\mu F\)
2 \(4\,\mu F\)
3 \(8\,\mu F\)
4 \(16\,\mu F\)
Explanation:
Let the arrangement be shown as and \({\therefore \quad \dfrac{1}{C_{\text {eq }}}=\dfrac{1}{C}+\dfrac{1}{C}}\) \({\therefore C_{\text {eq }}=C+C}\) \({C_{e q}=\dfrac{C^{2}}{2 C}}\) \({16 \mu {F}=2 {C}}\) \({\Rightarrow 4 \mu {F}=\dfrac{{C}^{2}}{2 C}}\) \({\therefore {C}=8 \mu {F}}\) \({\Rightarrow {C}=8 \mu {F}}\). So correct option is (3)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359288
All capacitors used in the diagram are identical and each is of capacitance \(C\). Then the effective capacitance between the points \(A\) and \(B\) is
1 \(3C\)
2 \(6C\)
3 \(1.5C\)
4 \(C\)
Explanation:
Let \(P\) be a point. The points having identical potentials are shown in the figure. The circuit can be redrawn as \({C_{AB}} = 1.5C.\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359289
The equivalent capacitance between points \({A}\) and \({B}\) of circuit shown in the figure is
1 \({9 \mu {F}}\)
2 \({1 \mu {F}}\)
3 \({4.5 \mu {F}}\)
4 \({6 \mu {F}}\)
Explanation:
Since, it is a parallel combination of all three capacitors. \({C_{eq}} = C + C + C = 3C = 3 \times 3\mu F,{C_{eq}} = 9\mu F.\) So correct option is (1)
359286
In figure below, four parallel plates of equal area \(A\) and spacing \(d\) are arrange, then effective capacitance between \(a\) and \(b\) is
1 \(\dfrac{\varepsilon_{0} A}{d}\)
2 \(\dfrac{2 \varepsilon_{0} A}{d}\)
3 \(\dfrac{3 \varepsilon_{0} A}{d}\)
4 \(\dfrac{4 \varepsilon_{0} A}{d}\)
Explanation:
Suppose the pair of plates \(a\) is connected to positive terminal of the battery and the pair of plate \(b\) is connected to negative terminal of the battery. Thus, from figure, we can conclude that, we have three capacitors I, II and III.
Their positive plates are connected to each other and so are the negative plates. Therefore, these are joined in parallel,\({\text{ }}i.e\). \(C_{p}=C_{1}+C_{2}+C_{3}=3 \varepsilon_{0} A / d\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359287
When two identical capacitors are in series, they have \(4\,\mu F\) capacitance and when in parallel \(16\,\mu F\). What is the capacitance of each?
1 \(6\,\mu F\)
2 \(4\,\mu F\)
3 \(8\,\mu F\)
4 \(16\,\mu F\)
Explanation:
Let the arrangement be shown as and \({\therefore \quad \dfrac{1}{C_{\text {eq }}}=\dfrac{1}{C}+\dfrac{1}{C}}\) \({\therefore C_{\text {eq }}=C+C}\) \({C_{e q}=\dfrac{C^{2}}{2 C}}\) \({16 \mu {F}=2 {C}}\) \({\Rightarrow 4 \mu {F}=\dfrac{{C}^{2}}{2 C}}\) \({\therefore {C}=8 \mu {F}}\) \({\Rightarrow {C}=8 \mu {F}}\). So correct option is (3)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359288
All capacitors used in the diagram are identical and each is of capacitance \(C\). Then the effective capacitance between the points \(A\) and \(B\) is
1 \(3C\)
2 \(6C\)
3 \(1.5C\)
4 \(C\)
Explanation:
Let \(P\) be a point. The points having identical potentials are shown in the figure. The circuit can be redrawn as \({C_{AB}} = 1.5C.\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359289
The equivalent capacitance between points \({A}\) and \({B}\) of circuit shown in the figure is
1 \({9 \mu {F}}\)
2 \({1 \mu {F}}\)
3 \({4.5 \mu {F}}\)
4 \({6 \mu {F}}\)
Explanation:
Since, it is a parallel combination of all three capacitors. \({C_{eq}} = C + C + C = 3C = 3 \times 3\mu F,{C_{eq}} = 9\mu F.\) So correct option is (1)
359286
In figure below, four parallel plates of equal area \(A\) and spacing \(d\) are arrange, then effective capacitance between \(a\) and \(b\) is
1 \(\dfrac{\varepsilon_{0} A}{d}\)
2 \(\dfrac{2 \varepsilon_{0} A}{d}\)
3 \(\dfrac{3 \varepsilon_{0} A}{d}\)
4 \(\dfrac{4 \varepsilon_{0} A}{d}\)
Explanation:
Suppose the pair of plates \(a\) is connected to positive terminal of the battery and the pair of plate \(b\) is connected to negative terminal of the battery. Thus, from figure, we can conclude that, we have three capacitors I, II and III.
Their positive plates are connected to each other and so are the negative plates. Therefore, these are joined in parallel,\({\text{ }}i.e\). \(C_{p}=C_{1}+C_{2}+C_{3}=3 \varepsilon_{0} A / d\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359287
When two identical capacitors are in series, they have \(4\,\mu F\) capacitance and when in parallel \(16\,\mu F\). What is the capacitance of each?
1 \(6\,\mu F\)
2 \(4\,\mu F\)
3 \(8\,\mu F\)
4 \(16\,\mu F\)
Explanation:
Let the arrangement be shown as and \({\therefore \quad \dfrac{1}{C_{\text {eq }}}=\dfrac{1}{C}+\dfrac{1}{C}}\) \({\therefore C_{\text {eq }}=C+C}\) \({C_{e q}=\dfrac{C^{2}}{2 C}}\) \({16 \mu {F}=2 {C}}\) \({\Rightarrow 4 \mu {F}=\dfrac{{C}^{2}}{2 C}}\) \({\therefore {C}=8 \mu {F}}\) \({\Rightarrow {C}=8 \mu {F}}\). So correct option is (3)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359288
All capacitors used in the diagram are identical and each is of capacitance \(C\). Then the effective capacitance between the points \(A\) and \(B\) is
1 \(3C\)
2 \(6C\)
3 \(1.5C\)
4 \(C\)
Explanation:
Let \(P\) be a point. The points having identical potentials are shown in the figure. The circuit can be redrawn as \({C_{AB}} = 1.5C.\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359289
The equivalent capacitance between points \({A}\) and \({B}\) of circuit shown in the figure is
1 \({9 \mu {F}}\)
2 \({1 \mu {F}}\)
3 \({4.5 \mu {F}}\)
4 \({6 \mu {F}}\)
Explanation:
Since, it is a parallel combination of all three capacitors. \({C_{eq}} = C + C + C = 3C = 3 \times 3\mu F,{C_{eq}} = 9\mu F.\) So correct option is (1)
