357499
If the condenser shown in the circuit is charged to \(5\;V\) and left in the circuit, in \(12\,\sec \) the charge on the condenser will become - \((e=2.718)\)
357500
Find the ratio of charges on \({6 \mu {F}}\) and \({4 \mu {F}}\) capacitors in the circuit shown in figure in steady state \({\left(\dfrac{q_{6 \mu {F}}}{q_{4 \mu {F}}}=?\right)}\).
1 0.15
2 0.61
3 0.34
4 0.75
Explanation:
Figure shows the potential distribution in the circuit as in steady state no current flows anywhere in the loop so consider resistances as short circuit for this case. We now write nodal equation for \(x\),\(3(x-15)+6(x-10)+2 x+4(x-5-0)=0\)\(15 x=125 \Rightarrow x=\dfrac{25}{3} {~V}\) Charge on \(6 \mu {F}\) capacitor, \(q_{6 \mu {F}}=6\left(10-\dfrac{25}{3}\right)=10 \mu {C}\) Charge on \(4 \mu {F}\) capacitor, \(q_{4 \mu {F}}=4\left(\dfrac{25}{3}-5\right)=\dfrac{40}{3} \mu {C}\) Hence required ratio, \(\dfrac{q_{6 \mu {F}}}{q_{4 \mu {F}}}=\dfrac{10}{40 / 3}=0.75\)
PHXII03:CURRENT ELECTRICITY
357501
In the circuit shown in Fig.\({C_1} = 2{C_2}.\) Intially, capacitor \({C_1}\) is charged to a potential of \(V\). The current in the circuit just after the switch is closed is
1 \(0\)
2 \(2V/R\)
3 \(\infty \)
4 \(V/2R\)
Explanation:
Uncharged capacitor \({C_2}\) behaves as a short circuit just after closing the switch. But charged capacitor \({C_1}\) behaves as the battery of emf V just after closing the switch. Current through the circuit is \(I = \frac{{{V_0}}}{{\left( {2R} \right)}} = \frac{{{q_0}}}{{2R{C_1}}} = \frac{V}{{2R}}\)
PHXII03:CURRENT ELECTRICITY
357502
In this figure, the resistance of the coil of galvanometer \(G\) is \(2 \Omega\). The emf of the cell is \(4\;V\). The ratio of potential difference across \(C_{1}\) and \(C_{2}\) is
1 \(\dfrac{4}{5}\)
2 \(\dfrac{3}{4}\)
3 \(\dfrac{5}{4}\)
4 1
Explanation:
Resistance of galvanometer \(=2 \Omega\) Equivalent resistance of coil \(=6+2+8=16 \Omega\) Current, \(i = \frac{4}{{16}} = \frac{1}{4}\;A\) In steady state, both the capacitors will act as open circuits \(\therefore V_{C_{1}}=i(6+2)=8 i\) \(\therefore {V_{{C_2}}} = i(8 + 2) = 10\,i\) \(\frac{{{V_{{C_1}}}}}{{{V_{{C_2}}}}} = \frac{8}{{10}} = \frac{4}{5}\)
357499
If the condenser shown in the circuit is charged to \(5\;V\) and left in the circuit, in \(12\,\sec \) the charge on the condenser will become - \((e=2.718)\)
357500
Find the ratio of charges on \({6 \mu {F}}\) and \({4 \mu {F}}\) capacitors in the circuit shown in figure in steady state \({\left(\dfrac{q_{6 \mu {F}}}{q_{4 \mu {F}}}=?\right)}\).
1 0.15
2 0.61
3 0.34
4 0.75
Explanation:
Figure shows the potential distribution in the circuit as in steady state no current flows anywhere in the loop so consider resistances as short circuit for this case. We now write nodal equation for \(x\),\(3(x-15)+6(x-10)+2 x+4(x-5-0)=0\)\(15 x=125 \Rightarrow x=\dfrac{25}{3} {~V}\) Charge on \(6 \mu {F}\) capacitor, \(q_{6 \mu {F}}=6\left(10-\dfrac{25}{3}\right)=10 \mu {C}\) Charge on \(4 \mu {F}\) capacitor, \(q_{4 \mu {F}}=4\left(\dfrac{25}{3}-5\right)=\dfrac{40}{3} \mu {C}\) Hence required ratio, \(\dfrac{q_{6 \mu {F}}}{q_{4 \mu {F}}}=\dfrac{10}{40 / 3}=0.75\)
PHXII03:CURRENT ELECTRICITY
357501
In the circuit shown in Fig.\({C_1} = 2{C_2}.\) Intially, capacitor \({C_1}\) is charged to a potential of \(V\). The current in the circuit just after the switch is closed is
1 \(0\)
2 \(2V/R\)
3 \(\infty \)
4 \(V/2R\)
Explanation:
Uncharged capacitor \({C_2}\) behaves as a short circuit just after closing the switch. But charged capacitor \({C_1}\) behaves as the battery of emf V just after closing the switch. Current through the circuit is \(I = \frac{{{V_0}}}{{\left( {2R} \right)}} = \frac{{{q_0}}}{{2R{C_1}}} = \frac{V}{{2R}}\)
PHXII03:CURRENT ELECTRICITY
357502
In this figure, the resistance of the coil of galvanometer \(G\) is \(2 \Omega\). The emf of the cell is \(4\;V\). The ratio of potential difference across \(C_{1}\) and \(C_{2}\) is
1 \(\dfrac{4}{5}\)
2 \(\dfrac{3}{4}\)
3 \(\dfrac{5}{4}\)
4 1
Explanation:
Resistance of galvanometer \(=2 \Omega\) Equivalent resistance of coil \(=6+2+8=16 \Omega\) Current, \(i = \frac{4}{{16}} = \frac{1}{4}\;A\) In steady state, both the capacitors will act as open circuits \(\therefore V_{C_{1}}=i(6+2)=8 i\) \(\therefore {V_{{C_2}}} = i(8 + 2) = 10\,i\) \(\frac{{{V_{{C_1}}}}}{{{V_{{C_2}}}}} = \frac{8}{{10}} = \frac{4}{5}\)
357499
If the condenser shown in the circuit is charged to \(5\;V\) and left in the circuit, in \(12\,\sec \) the charge on the condenser will become - \((e=2.718)\)
357500
Find the ratio of charges on \({6 \mu {F}}\) and \({4 \mu {F}}\) capacitors in the circuit shown in figure in steady state \({\left(\dfrac{q_{6 \mu {F}}}{q_{4 \mu {F}}}=?\right)}\).
1 0.15
2 0.61
3 0.34
4 0.75
Explanation:
Figure shows the potential distribution in the circuit as in steady state no current flows anywhere in the loop so consider resistances as short circuit for this case. We now write nodal equation for \(x\),\(3(x-15)+6(x-10)+2 x+4(x-5-0)=0\)\(15 x=125 \Rightarrow x=\dfrac{25}{3} {~V}\) Charge on \(6 \mu {F}\) capacitor, \(q_{6 \mu {F}}=6\left(10-\dfrac{25}{3}\right)=10 \mu {C}\) Charge on \(4 \mu {F}\) capacitor, \(q_{4 \mu {F}}=4\left(\dfrac{25}{3}-5\right)=\dfrac{40}{3} \mu {C}\) Hence required ratio, \(\dfrac{q_{6 \mu {F}}}{q_{4 \mu {F}}}=\dfrac{10}{40 / 3}=0.75\)
PHXII03:CURRENT ELECTRICITY
357501
In the circuit shown in Fig.\({C_1} = 2{C_2}.\) Intially, capacitor \({C_1}\) is charged to a potential of \(V\). The current in the circuit just after the switch is closed is
1 \(0\)
2 \(2V/R\)
3 \(\infty \)
4 \(V/2R\)
Explanation:
Uncharged capacitor \({C_2}\) behaves as a short circuit just after closing the switch. But charged capacitor \({C_1}\) behaves as the battery of emf V just after closing the switch. Current through the circuit is \(I = \frac{{{V_0}}}{{\left( {2R} \right)}} = \frac{{{q_0}}}{{2R{C_1}}} = \frac{V}{{2R}}\)
PHXII03:CURRENT ELECTRICITY
357502
In this figure, the resistance of the coil of galvanometer \(G\) is \(2 \Omega\). The emf of the cell is \(4\;V\). The ratio of potential difference across \(C_{1}\) and \(C_{2}\) is
1 \(\dfrac{4}{5}\)
2 \(\dfrac{3}{4}\)
3 \(\dfrac{5}{4}\)
4 1
Explanation:
Resistance of galvanometer \(=2 \Omega\) Equivalent resistance of coil \(=6+2+8=16 \Omega\) Current, \(i = \frac{4}{{16}} = \frac{1}{4}\;A\) In steady state, both the capacitors will act as open circuits \(\therefore V_{C_{1}}=i(6+2)=8 i\) \(\therefore {V_{{C_2}}} = i(8 + 2) = 10\,i\) \(\frac{{{V_{{C_1}}}}}{{{V_{{C_2}}}}} = \frac{8}{{10}} = \frac{4}{5}\)
357499
If the condenser shown in the circuit is charged to \(5\;V\) and left in the circuit, in \(12\,\sec \) the charge on the condenser will become - \((e=2.718)\)
357500
Find the ratio of charges on \({6 \mu {F}}\) and \({4 \mu {F}}\) capacitors in the circuit shown in figure in steady state \({\left(\dfrac{q_{6 \mu {F}}}{q_{4 \mu {F}}}=?\right)}\).
1 0.15
2 0.61
3 0.34
4 0.75
Explanation:
Figure shows the potential distribution in the circuit as in steady state no current flows anywhere in the loop so consider resistances as short circuit for this case. We now write nodal equation for \(x\),\(3(x-15)+6(x-10)+2 x+4(x-5-0)=0\)\(15 x=125 \Rightarrow x=\dfrac{25}{3} {~V}\) Charge on \(6 \mu {F}\) capacitor, \(q_{6 \mu {F}}=6\left(10-\dfrac{25}{3}\right)=10 \mu {C}\) Charge on \(4 \mu {F}\) capacitor, \(q_{4 \mu {F}}=4\left(\dfrac{25}{3}-5\right)=\dfrac{40}{3} \mu {C}\) Hence required ratio, \(\dfrac{q_{6 \mu {F}}}{q_{4 \mu {F}}}=\dfrac{10}{40 / 3}=0.75\)
PHXII03:CURRENT ELECTRICITY
357501
In the circuit shown in Fig.\({C_1} = 2{C_2}.\) Intially, capacitor \({C_1}\) is charged to a potential of \(V\). The current in the circuit just after the switch is closed is
1 \(0\)
2 \(2V/R\)
3 \(\infty \)
4 \(V/2R\)
Explanation:
Uncharged capacitor \({C_2}\) behaves as a short circuit just after closing the switch. But charged capacitor \({C_1}\) behaves as the battery of emf V just after closing the switch. Current through the circuit is \(I = \frac{{{V_0}}}{{\left( {2R} \right)}} = \frac{{{q_0}}}{{2R{C_1}}} = \frac{V}{{2R}}\)
PHXII03:CURRENT ELECTRICITY
357502
In this figure, the resistance of the coil of galvanometer \(G\) is \(2 \Omega\). The emf of the cell is \(4\;V\). The ratio of potential difference across \(C_{1}\) and \(C_{2}\) is
1 \(\dfrac{4}{5}\)
2 \(\dfrac{3}{4}\)
3 \(\dfrac{5}{4}\)
4 1
Explanation:
Resistance of galvanometer \(=2 \Omega\) Equivalent resistance of coil \(=6+2+8=16 \Omega\) Current, \(i = \frac{4}{{16}} = \frac{1}{4}\;A\) In steady state, both the capacitors will act as open circuits \(\therefore V_{C_{1}}=i(6+2)=8 i\) \(\therefore {V_{{C_2}}} = i(8 + 2) = 10\,i\) \(\frac{{{V_{{C_1}}}}}{{{V_{{C_2}}}}} = \frac{8}{{10}} = \frac{4}{5}\)
