359277
Five identical plates are connected across a battery as follows. If the charge on plate 1 be \(+q\), then the charges on the plates 2,3,4 and 5 are
1 \( - q, + q, + q\)
2 \( - 2q, + 2q, - 2q, + q\)
3 \( - q, + 2q, - 2q, + q\)
4 None of the above
Explanation:
From the given figure plates 1,3 and 5 are connected to (+) \(ve\) terminal of the battery. Plates 2 and 4 are connected (-) ve terminal of the battery. The capacitance of all five capacitors are the same. The potential difference between any two adjacent plates is \(V\). The charge on all five capacitors are the same. \(q = \frac{{{\varepsilon _0}AV}}{d}\) charges on the plates 2,3,4 and 5 are \({q_2} = - 2q,\,\,{q_3} = 2q,\;{q_4} = - 2q\;and\,{q_5} = q\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359278
In the given network, the value of \(C\), so the an equivalent capacitance between \(A\) and \(B\) is \(3\mu F\), is
1 \(\frac{1}{5}\mu F\)
2 \(\frac{{31}}{5}\mu F\)
3 \(48\mu F\)
4 \(36\mu F\)
Explanation:
The equivalent capacitance between \(A\) and \(B\) is \(\frac{1}{{{C_{AB}}}}\; = \;\frac{1}{C}\; + \,\frac{1}{{\frac{{16}}{5}}}\) \(\frac{1}{C}\; = \,\frac{1}{{{C_{AB}}}}\, - \;\frac{5}{{16}}\; = \,\frac{1}{3}\; - \,\frac{5}{{16}}\, = \;\frac{{16\; - \;15}}{{48}}\; = \;\frac{1}{{48}}\) \(C = 48\mu F\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359279
Six capacitors each having a capacity of \(2\mu F\) are connected as shown in the figure. The effective capacitance between \(A\) and \(B\) is
1 \(6\mu F\)
2 \(3\mu F\)
3 \(12\mu F\)
4 \(\frac{2}{3}\mu F\)
Explanation:
All the capacitors are connected between \(A\) and \(B\) and hence given six capacitors are in parallel. \(\because {C_{eq}} = 6C = 6 \times 2\mu F = 12\mu F\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359280
Three capacitances, each of \(3\;\mu F,\) are provided. These cannot be combined to provide the resultant capacitance of :
1 \(4.5\;\mu F\)
2 \(1\;\mu F\)
3 \(2\;\mu F\)
4 \(6\;\mu F\)
Explanation:
When all are in series \({C_{eq}} = 1\;\mu F\) When all are in parallel \({C_{eq}} = 9\;\mu F\) When two are in parallel and one in series \({C_{eq}} = \frac{{6 \times 3}}{9} = 2\;\mu F\) When two are in series and one is in parallel \(\frac{{3 \times 3}}{6} + 3 = 4.5\;\mu F\) Hence answer is (4)
359277
Five identical plates are connected across a battery as follows. If the charge on plate 1 be \(+q\), then the charges on the plates 2,3,4 and 5 are
1 \( - q, + q, + q\)
2 \( - 2q, + 2q, - 2q, + q\)
3 \( - q, + 2q, - 2q, + q\)
4 None of the above
Explanation:
From the given figure plates 1,3 and 5 are connected to (+) \(ve\) terminal of the battery. Plates 2 and 4 are connected (-) ve terminal of the battery. The capacitance of all five capacitors are the same. The potential difference between any two adjacent plates is \(V\). The charge on all five capacitors are the same. \(q = \frac{{{\varepsilon _0}AV}}{d}\) charges on the plates 2,3,4 and 5 are \({q_2} = - 2q,\,\,{q_3} = 2q,\;{q_4} = - 2q\;and\,{q_5} = q\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359278
In the given network, the value of \(C\), so the an equivalent capacitance between \(A\) and \(B\) is \(3\mu F\), is
1 \(\frac{1}{5}\mu F\)
2 \(\frac{{31}}{5}\mu F\)
3 \(48\mu F\)
4 \(36\mu F\)
Explanation:
The equivalent capacitance between \(A\) and \(B\) is \(\frac{1}{{{C_{AB}}}}\; = \;\frac{1}{C}\; + \,\frac{1}{{\frac{{16}}{5}}}\) \(\frac{1}{C}\; = \,\frac{1}{{{C_{AB}}}}\, - \;\frac{5}{{16}}\; = \,\frac{1}{3}\; - \,\frac{5}{{16}}\, = \;\frac{{16\; - \;15}}{{48}}\; = \;\frac{1}{{48}}\) \(C = 48\mu F\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359279
Six capacitors each having a capacity of \(2\mu F\) are connected as shown in the figure. The effective capacitance between \(A\) and \(B\) is
1 \(6\mu F\)
2 \(3\mu F\)
3 \(12\mu F\)
4 \(\frac{2}{3}\mu F\)
Explanation:
All the capacitors are connected between \(A\) and \(B\) and hence given six capacitors are in parallel. \(\because {C_{eq}} = 6C = 6 \times 2\mu F = 12\mu F\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359280
Three capacitances, each of \(3\;\mu F,\) are provided. These cannot be combined to provide the resultant capacitance of :
1 \(4.5\;\mu F\)
2 \(1\;\mu F\)
3 \(2\;\mu F\)
4 \(6\;\mu F\)
Explanation:
When all are in series \({C_{eq}} = 1\;\mu F\) When all are in parallel \({C_{eq}} = 9\;\mu F\) When two are in parallel and one in series \({C_{eq}} = \frac{{6 \times 3}}{9} = 2\;\mu F\) When two are in series and one is in parallel \(\frac{{3 \times 3}}{6} + 3 = 4.5\;\mu F\) Hence answer is (4)
359277
Five identical plates are connected across a battery as follows. If the charge on plate 1 be \(+q\), then the charges on the plates 2,3,4 and 5 are
1 \( - q, + q, + q\)
2 \( - 2q, + 2q, - 2q, + q\)
3 \( - q, + 2q, - 2q, + q\)
4 None of the above
Explanation:
From the given figure plates 1,3 and 5 are connected to (+) \(ve\) terminal of the battery. Plates 2 and 4 are connected (-) ve terminal of the battery. The capacitance of all five capacitors are the same. The potential difference between any two adjacent plates is \(V\). The charge on all five capacitors are the same. \(q = \frac{{{\varepsilon _0}AV}}{d}\) charges on the plates 2,3,4 and 5 are \({q_2} = - 2q,\,\,{q_3} = 2q,\;{q_4} = - 2q\;and\,{q_5} = q\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359278
In the given network, the value of \(C\), so the an equivalent capacitance between \(A\) and \(B\) is \(3\mu F\), is
1 \(\frac{1}{5}\mu F\)
2 \(\frac{{31}}{5}\mu F\)
3 \(48\mu F\)
4 \(36\mu F\)
Explanation:
The equivalent capacitance between \(A\) and \(B\) is \(\frac{1}{{{C_{AB}}}}\; = \;\frac{1}{C}\; + \,\frac{1}{{\frac{{16}}{5}}}\) \(\frac{1}{C}\; = \,\frac{1}{{{C_{AB}}}}\, - \;\frac{5}{{16}}\; = \,\frac{1}{3}\; - \,\frac{5}{{16}}\, = \;\frac{{16\; - \;15}}{{48}}\; = \;\frac{1}{{48}}\) \(C = 48\mu F\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359279
Six capacitors each having a capacity of \(2\mu F\) are connected as shown in the figure. The effective capacitance between \(A\) and \(B\) is
1 \(6\mu F\)
2 \(3\mu F\)
3 \(12\mu F\)
4 \(\frac{2}{3}\mu F\)
Explanation:
All the capacitors are connected between \(A\) and \(B\) and hence given six capacitors are in parallel. \(\because {C_{eq}} = 6C = 6 \times 2\mu F = 12\mu F\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359280
Three capacitances, each of \(3\;\mu F,\) are provided. These cannot be combined to provide the resultant capacitance of :
1 \(4.5\;\mu F\)
2 \(1\;\mu F\)
3 \(2\;\mu F\)
4 \(6\;\mu F\)
Explanation:
When all are in series \({C_{eq}} = 1\;\mu F\) When all are in parallel \({C_{eq}} = 9\;\mu F\) When two are in parallel and one in series \({C_{eq}} = \frac{{6 \times 3}}{9} = 2\;\mu F\) When two are in series and one is in parallel \(\frac{{3 \times 3}}{6} + 3 = 4.5\;\mu F\) Hence answer is (4)
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PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359277
Five identical plates are connected across a battery as follows. If the charge on plate 1 be \(+q\), then the charges on the plates 2,3,4 and 5 are
1 \( - q, + q, + q\)
2 \( - 2q, + 2q, - 2q, + q\)
3 \( - q, + 2q, - 2q, + q\)
4 None of the above
Explanation:
From the given figure plates 1,3 and 5 are connected to (+) \(ve\) terminal of the battery. Plates 2 and 4 are connected (-) ve terminal of the battery. The capacitance of all five capacitors are the same. The potential difference between any two adjacent plates is \(V\). The charge on all five capacitors are the same. \(q = \frac{{{\varepsilon _0}AV}}{d}\) charges on the plates 2,3,4 and 5 are \({q_2} = - 2q,\,\,{q_3} = 2q,\;{q_4} = - 2q\;and\,{q_5} = q\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359278
In the given network, the value of \(C\), so the an equivalent capacitance between \(A\) and \(B\) is \(3\mu F\), is
1 \(\frac{1}{5}\mu F\)
2 \(\frac{{31}}{5}\mu F\)
3 \(48\mu F\)
4 \(36\mu F\)
Explanation:
The equivalent capacitance between \(A\) and \(B\) is \(\frac{1}{{{C_{AB}}}}\; = \;\frac{1}{C}\; + \,\frac{1}{{\frac{{16}}{5}}}\) \(\frac{1}{C}\; = \,\frac{1}{{{C_{AB}}}}\, - \;\frac{5}{{16}}\; = \,\frac{1}{3}\; - \,\frac{5}{{16}}\, = \;\frac{{16\; - \;15}}{{48}}\; = \;\frac{1}{{48}}\) \(C = 48\mu F\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359279
Six capacitors each having a capacity of \(2\mu F\) are connected as shown in the figure. The effective capacitance between \(A\) and \(B\) is
1 \(6\mu F\)
2 \(3\mu F\)
3 \(12\mu F\)
4 \(\frac{2}{3}\mu F\)
Explanation:
All the capacitors are connected between \(A\) and \(B\) and hence given six capacitors are in parallel. \(\because {C_{eq}} = 6C = 6 \times 2\mu F = 12\mu F\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359280
Three capacitances, each of \(3\;\mu F,\) are provided. These cannot be combined to provide the resultant capacitance of :
1 \(4.5\;\mu F\)
2 \(1\;\mu F\)
3 \(2\;\mu F\)
4 \(6\;\mu F\)
Explanation:
When all are in series \({C_{eq}} = 1\;\mu F\) When all are in parallel \({C_{eq}} = 9\;\mu F\) When two are in parallel and one in series \({C_{eq}} = \frac{{6 \times 3}}{9} = 2\;\mu F\) When two are in series and one is in parallel \(\frac{{3 \times 3}}{6} + 3 = 4.5\;\mu F\) Hence answer is (4)
