359243
Four equal capacitors, each of capacity \(C\), are arranged as shown. The effective capacitance between \(A\) and \(B\) is
1 \(C\)
2 \(\frac{5}{3}C\)
3 \(\frac{5}{8}C\)
4 \(\frac{3}{5}C\)
Explanation:
The given circuit can be simplified as follows:
Equivalent capacitance between \(A\) and \(B\) is \({C_{AB}} = \frac{5}{3}C\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359244
In the given network capacitance \({C_2} = 10\,\mu F,\,\,{C_1} = 5\,\mu {\rm{F}}\) and \({C_3} = 4\,\mu F\). The resultant capacitance between \(P\) and \(Q\) will be:
1 \(4.7 \mu \mathrm{F}\)
2 \(1.2 \mu F\)
3 \(3.2 \mu F\)
4 \(2.2 \mu F\)
Explanation:
Seeing in the given circuit \(C_{1}\) and \(C_{2}\) are connected in parallel. Hence, their equivalent capacitance \(C_{e q}=C_{1}+C_{2}=5+10=15 \mu F\) As \(C_{e q}\) and \(C_{3}\) are connected in series, Hence, resultant capacitance between \(P\) and \(Q\) is given by \(\dfrac{1}{C_{P Q}}=\dfrac{1}{C_{e q}}+\dfrac{1}{C_{3}}=\dfrac{1}{15}+\dfrac{1}{4}=\dfrac{19}{60}\) \(\Rightarrow C_{P Q}=\dfrac{60}{19}=3.2 \mu F\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359245
When two identical capacitors are in series have \(3\mu F\;\) capacitance and when parallel \(12\mu F\;\). What is the capacitance of each capacitor?
1 \(6\mu F\;\)
2 \(3\mu F\;\)
3 \(12\mu F\;\)
4 \(9\mu F\;\)
Explanation:
For series combination \(\frac{C}{2} = 3\mu F \Rightarrow C = 6\mu F\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359246
Five identical capacitor plates each of area \(A\) are arranged such that adjacent plates are at a distance d apart. The plates are connected to a source of emf \(V\) as shown in figure. What is the charge on plate?
1 \(\frac{{{\varepsilon _0}AV}}{d}\)
2 \(\frac{{{\varepsilon _0}AV}}{{2d}}\)
3 \(\frac{{2{\varepsilon _0}AV}}{d}\)
4 \(\frac{{{\varepsilon _0}AV}}{{4d}}\)
Explanation:
Plate 1 and 2 forms a capacitor and the potential difference across them is \(V\). The charge on the capacitor (Plate -1) is \(q = CV = \frac{{{\varepsilon _o}AV}}{d}\)
359243
Four equal capacitors, each of capacity \(C\), are arranged as shown. The effective capacitance between \(A\) and \(B\) is
1 \(C\)
2 \(\frac{5}{3}C\)
3 \(\frac{5}{8}C\)
4 \(\frac{3}{5}C\)
Explanation:
The given circuit can be simplified as follows:
Equivalent capacitance between \(A\) and \(B\) is \({C_{AB}} = \frac{5}{3}C\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359244
In the given network capacitance \({C_2} = 10\,\mu F,\,\,{C_1} = 5\,\mu {\rm{F}}\) and \({C_3} = 4\,\mu F\). The resultant capacitance between \(P\) and \(Q\) will be:
1 \(4.7 \mu \mathrm{F}\)
2 \(1.2 \mu F\)
3 \(3.2 \mu F\)
4 \(2.2 \mu F\)
Explanation:
Seeing in the given circuit \(C_{1}\) and \(C_{2}\) are connected in parallel. Hence, their equivalent capacitance \(C_{e q}=C_{1}+C_{2}=5+10=15 \mu F\) As \(C_{e q}\) and \(C_{3}\) are connected in series, Hence, resultant capacitance between \(P\) and \(Q\) is given by \(\dfrac{1}{C_{P Q}}=\dfrac{1}{C_{e q}}+\dfrac{1}{C_{3}}=\dfrac{1}{15}+\dfrac{1}{4}=\dfrac{19}{60}\) \(\Rightarrow C_{P Q}=\dfrac{60}{19}=3.2 \mu F\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359245
When two identical capacitors are in series have \(3\mu F\;\) capacitance and when parallel \(12\mu F\;\). What is the capacitance of each capacitor?
1 \(6\mu F\;\)
2 \(3\mu F\;\)
3 \(12\mu F\;\)
4 \(9\mu F\;\)
Explanation:
For series combination \(\frac{C}{2} = 3\mu F \Rightarrow C = 6\mu F\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359246
Five identical capacitor plates each of area \(A\) are arranged such that adjacent plates are at a distance d apart. The plates are connected to a source of emf \(V\) as shown in figure. What is the charge on plate?
1 \(\frac{{{\varepsilon _0}AV}}{d}\)
2 \(\frac{{{\varepsilon _0}AV}}{{2d}}\)
3 \(\frac{{2{\varepsilon _0}AV}}{d}\)
4 \(\frac{{{\varepsilon _0}AV}}{{4d}}\)
Explanation:
Plate 1 and 2 forms a capacitor and the potential difference across them is \(V\). The charge on the capacitor (Plate -1) is \(q = CV = \frac{{{\varepsilon _o}AV}}{d}\)
359243
Four equal capacitors, each of capacity \(C\), are arranged as shown. The effective capacitance between \(A\) and \(B\) is
1 \(C\)
2 \(\frac{5}{3}C\)
3 \(\frac{5}{8}C\)
4 \(\frac{3}{5}C\)
Explanation:
The given circuit can be simplified as follows:
Equivalent capacitance between \(A\) and \(B\) is \({C_{AB}} = \frac{5}{3}C\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359244
In the given network capacitance \({C_2} = 10\,\mu F,\,\,{C_1} = 5\,\mu {\rm{F}}\) and \({C_3} = 4\,\mu F\). The resultant capacitance between \(P\) and \(Q\) will be:
1 \(4.7 \mu \mathrm{F}\)
2 \(1.2 \mu F\)
3 \(3.2 \mu F\)
4 \(2.2 \mu F\)
Explanation:
Seeing in the given circuit \(C_{1}\) and \(C_{2}\) are connected in parallel. Hence, their equivalent capacitance \(C_{e q}=C_{1}+C_{2}=5+10=15 \mu F\) As \(C_{e q}\) and \(C_{3}\) are connected in series, Hence, resultant capacitance between \(P\) and \(Q\) is given by \(\dfrac{1}{C_{P Q}}=\dfrac{1}{C_{e q}}+\dfrac{1}{C_{3}}=\dfrac{1}{15}+\dfrac{1}{4}=\dfrac{19}{60}\) \(\Rightarrow C_{P Q}=\dfrac{60}{19}=3.2 \mu F\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359245
When two identical capacitors are in series have \(3\mu F\;\) capacitance and when parallel \(12\mu F\;\). What is the capacitance of each capacitor?
1 \(6\mu F\;\)
2 \(3\mu F\;\)
3 \(12\mu F\;\)
4 \(9\mu F\;\)
Explanation:
For series combination \(\frac{C}{2} = 3\mu F \Rightarrow C = 6\mu F\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359246
Five identical capacitor plates each of area \(A\) are arranged such that adjacent plates are at a distance d apart. The plates are connected to a source of emf \(V\) as shown in figure. What is the charge on plate?
1 \(\frac{{{\varepsilon _0}AV}}{d}\)
2 \(\frac{{{\varepsilon _0}AV}}{{2d}}\)
3 \(\frac{{2{\varepsilon _0}AV}}{d}\)
4 \(\frac{{{\varepsilon _0}AV}}{{4d}}\)
Explanation:
Plate 1 and 2 forms a capacitor and the potential difference across them is \(V\). The charge on the capacitor (Plate -1) is \(q = CV = \frac{{{\varepsilon _o}AV}}{d}\)
359243
Four equal capacitors, each of capacity \(C\), are arranged as shown. The effective capacitance between \(A\) and \(B\) is
1 \(C\)
2 \(\frac{5}{3}C\)
3 \(\frac{5}{8}C\)
4 \(\frac{3}{5}C\)
Explanation:
The given circuit can be simplified as follows:
Equivalent capacitance between \(A\) and \(B\) is \({C_{AB}} = \frac{5}{3}C\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359244
In the given network capacitance \({C_2} = 10\,\mu F,\,\,{C_1} = 5\,\mu {\rm{F}}\) and \({C_3} = 4\,\mu F\). The resultant capacitance between \(P\) and \(Q\) will be:
1 \(4.7 \mu \mathrm{F}\)
2 \(1.2 \mu F\)
3 \(3.2 \mu F\)
4 \(2.2 \mu F\)
Explanation:
Seeing in the given circuit \(C_{1}\) and \(C_{2}\) are connected in parallel. Hence, their equivalent capacitance \(C_{e q}=C_{1}+C_{2}=5+10=15 \mu F\) As \(C_{e q}\) and \(C_{3}\) are connected in series, Hence, resultant capacitance between \(P\) and \(Q\) is given by \(\dfrac{1}{C_{P Q}}=\dfrac{1}{C_{e q}}+\dfrac{1}{C_{3}}=\dfrac{1}{15}+\dfrac{1}{4}=\dfrac{19}{60}\) \(\Rightarrow C_{P Q}=\dfrac{60}{19}=3.2 \mu F\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359245
When two identical capacitors are in series have \(3\mu F\;\) capacitance and when parallel \(12\mu F\;\). What is the capacitance of each capacitor?
1 \(6\mu F\;\)
2 \(3\mu F\;\)
3 \(12\mu F\;\)
4 \(9\mu F\;\)
Explanation:
For series combination \(\frac{C}{2} = 3\mu F \Rightarrow C = 6\mu F\)
PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE
359246
Five identical capacitor plates each of area \(A\) are arranged such that adjacent plates are at a distance d apart. The plates are connected to a source of emf \(V\) as shown in figure. What is the charge on plate?
1 \(\frac{{{\varepsilon _0}AV}}{d}\)
2 \(\frac{{{\varepsilon _0}AV}}{{2d}}\)
3 \(\frac{{2{\varepsilon _0}AV}}{d}\)
4 \(\frac{{{\varepsilon _0}AV}}{{4d}}\)
Explanation:
Plate 1 and 2 forms a capacitor and the potential difference across them is \(V\). The charge on the capacitor (Plate -1) is \(q = CV = \frac{{{\varepsilon _o}AV}}{d}\)
