In steady state, current through \(AB = 0\) \( \Rightarrow {V_{AB}} = {V_{CD}}\) The current through outermost loop is \(i = \frac{E}{{r + {r_2}}}\) \( \Rightarrow {V_{AB}} = \left( {\frac{E}{{r + {r_2}}} \times {r_2}} \right)\) As potential difference across \({r_1}\) is zero. \( \Rightarrow Q = C{V_{AB}} = CE\left( {\frac{{{r_2}}}{{r + {r_2}}}} \right)\)
PHXII03:CURRENT ELECTRICITY
357491
In the given circuit diagram the current through the battery and the charge on the capacitor respectively in steady state is
1 \(11A\,\,{\rm{and}}\,\,{\rm{3}}\mu C\)
2 \(1A\,\,{\rm{and}}\,\,{\rm{3}}\mu C\)
3 \(6A\,\,{\rm{and}}\,\,0\mu C\)
4 \(17A\,\,{\rm{and}}\,\,0\mu C\)
Explanation:
In steady state capacitor is fully charged and no current flows through it. \(\therefore \) No current passes through \(4\Omega \) \(\frac{1}{{{R_{eff}}}} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} = \frac{{6 + 3 + 2}}{6} = \frac{{11}}{6}\) \( \Rightarrow {R_{eff}} = \frac{6}{{11}}\Omega \) \({\rm{Current = }}\frac{{6 \times 11}}{6} = 11A\) \(Q = CV = 0.5 \times {10^{ - 6}} \times 6 = 3.0 \times {10^{ - 6}}C = 3\,\mu C\)
PHXII03:CURRENT ELECTRICITY
357492
Consider the circuit given below and calculate the charge on the capacitors.
1 \(5\,\mu C\)
2 \(1\,\mu C\)
3 \(3\,\mu C\)
4 \(7\,\mu C\)
Explanation:
\(I = \frac{3}{{6 \times {{10}^3}}} = 0.5 \times {10^{ - 3}}\;A\) \({V}_{{AD}}={IR}=0.5 \times 10^{-3} \times(2+1) \times 10^{3}=1.5 {~V}\) The equivalent capacitance of the given capacitors will be, \(\dfrac{1}{{C}_{{cq}}}=\dfrac{1}{{C}_{1}}+\dfrac{1}{{C}_{2}}=\dfrac{1}{1}+\dfrac{1}{2}=\dfrac{3}{2}\) \(\Rightarrow {C}_{{eq}}=\dfrac{2}{3} \mu {F}\) \(\therefore\) Charge on the capacitors, \({Q}=\dfrac{2 \times 10^{-6} \times 1.5}{3}=2 \times 10^{-6} \times 0.5=1 \mu {C}\)
PHXII03:CURRENT ELECTRICITY
357493
When the key \({K}\) is pressed at time \({t=0}\), which of the following is true about the current \({I}\) in the resistor \({A B}\) ?
1 2 \(mA\) all the time
2 oscillates between 1 \(mA\) and 2 \(mA\)
3 1 \(mA\) all the time
4 At \(t = 0\), \(I = 2\,\,mA\) and reduces to 1 \(mA\) finally
Explanation:
At \({t=0, I=\dfrac{2 {~V}}{1 {k} \Omega}=2 {~mA}}\) When capacitor is fully charged it will behave as open circuit, Net resistance of circuit, \({R=1 {k} \Omega+1 {k} \Omega=2 {k} \Omega}\) \(\therefore {\text{Current}} = \frac{{2\;V}}{{2\,k\,\Omega }} = 1\;\,mA\) Finally current will reduce to 1 mA . So correct option is (4)
PHXII03:CURRENT ELECTRICITY
357494
The circuit shown here has two batteries of 8.0\(V\) and 16.0 \(V\) and three resistors \(3\Omega \,,9\Omega \) and \(9\Omega \) and a capacitor \(5.0\,\mu F.\) How much is the current I in the circuit in steady state?
1 1.6 \(A\)
2 2.5 \(A\)
3 0.67 \(A\)
4 0.25 \(A\)
Explanation:
In the steady state current does not flow hrough the capacitor. The current in the outer circuit is \(16V - 9i - 3i - 8V = 0\) \(i = \frac{8}{{12}} = 0.67A\)
In steady state, current through \(AB = 0\) \( \Rightarrow {V_{AB}} = {V_{CD}}\) The current through outermost loop is \(i = \frac{E}{{r + {r_2}}}\) \( \Rightarrow {V_{AB}} = \left( {\frac{E}{{r + {r_2}}} \times {r_2}} \right)\) As potential difference across \({r_1}\) is zero. \( \Rightarrow Q = C{V_{AB}} = CE\left( {\frac{{{r_2}}}{{r + {r_2}}}} \right)\)
PHXII03:CURRENT ELECTRICITY
357491
In the given circuit diagram the current through the battery and the charge on the capacitor respectively in steady state is
1 \(11A\,\,{\rm{and}}\,\,{\rm{3}}\mu C\)
2 \(1A\,\,{\rm{and}}\,\,{\rm{3}}\mu C\)
3 \(6A\,\,{\rm{and}}\,\,0\mu C\)
4 \(17A\,\,{\rm{and}}\,\,0\mu C\)
Explanation:
In steady state capacitor is fully charged and no current flows through it. \(\therefore \) No current passes through \(4\Omega \) \(\frac{1}{{{R_{eff}}}} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} = \frac{{6 + 3 + 2}}{6} = \frac{{11}}{6}\) \( \Rightarrow {R_{eff}} = \frac{6}{{11}}\Omega \) \({\rm{Current = }}\frac{{6 \times 11}}{6} = 11A\) \(Q = CV = 0.5 \times {10^{ - 6}} \times 6 = 3.0 \times {10^{ - 6}}C = 3\,\mu C\)
PHXII03:CURRENT ELECTRICITY
357492
Consider the circuit given below and calculate the charge on the capacitors.
1 \(5\,\mu C\)
2 \(1\,\mu C\)
3 \(3\,\mu C\)
4 \(7\,\mu C\)
Explanation:
\(I = \frac{3}{{6 \times {{10}^3}}} = 0.5 \times {10^{ - 3}}\;A\) \({V}_{{AD}}={IR}=0.5 \times 10^{-3} \times(2+1) \times 10^{3}=1.5 {~V}\) The equivalent capacitance of the given capacitors will be, \(\dfrac{1}{{C}_{{cq}}}=\dfrac{1}{{C}_{1}}+\dfrac{1}{{C}_{2}}=\dfrac{1}{1}+\dfrac{1}{2}=\dfrac{3}{2}\) \(\Rightarrow {C}_{{eq}}=\dfrac{2}{3} \mu {F}\) \(\therefore\) Charge on the capacitors, \({Q}=\dfrac{2 \times 10^{-6} \times 1.5}{3}=2 \times 10^{-6} \times 0.5=1 \mu {C}\)
PHXII03:CURRENT ELECTRICITY
357493
When the key \({K}\) is pressed at time \({t=0}\), which of the following is true about the current \({I}\) in the resistor \({A B}\) ?
1 2 \(mA\) all the time
2 oscillates between 1 \(mA\) and 2 \(mA\)
3 1 \(mA\) all the time
4 At \(t = 0\), \(I = 2\,\,mA\) and reduces to 1 \(mA\) finally
Explanation:
At \({t=0, I=\dfrac{2 {~V}}{1 {k} \Omega}=2 {~mA}}\) When capacitor is fully charged it will behave as open circuit, Net resistance of circuit, \({R=1 {k} \Omega+1 {k} \Omega=2 {k} \Omega}\) \(\therefore {\text{Current}} = \frac{{2\;V}}{{2\,k\,\Omega }} = 1\;\,mA\) Finally current will reduce to 1 mA . So correct option is (4)
PHXII03:CURRENT ELECTRICITY
357494
The circuit shown here has two batteries of 8.0\(V\) and 16.0 \(V\) and three resistors \(3\Omega \,,9\Omega \) and \(9\Omega \) and a capacitor \(5.0\,\mu F.\) How much is the current I in the circuit in steady state?
1 1.6 \(A\)
2 2.5 \(A\)
3 0.67 \(A\)
4 0.25 \(A\)
Explanation:
In the steady state current does not flow hrough the capacitor. The current in the outer circuit is \(16V - 9i - 3i - 8V = 0\) \(i = \frac{8}{{12}} = 0.67A\)
In steady state, current through \(AB = 0\) \( \Rightarrow {V_{AB}} = {V_{CD}}\) The current through outermost loop is \(i = \frac{E}{{r + {r_2}}}\) \( \Rightarrow {V_{AB}} = \left( {\frac{E}{{r + {r_2}}} \times {r_2}} \right)\) As potential difference across \({r_1}\) is zero. \( \Rightarrow Q = C{V_{AB}} = CE\left( {\frac{{{r_2}}}{{r + {r_2}}}} \right)\)
PHXII03:CURRENT ELECTRICITY
357491
In the given circuit diagram the current through the battery and the charge on the capacitor respectively in steady state is
1 \(11A\,\,{\rm{and}}\,\,{\rm{3}}\mu C\)
2 \(1A\,\,{\rm{and}}\,\,{\rm{3}}\mu C\)
3 \(6A\,\,{\rm{and}}\,\,0\mu C\)
4 \(17A\,\,{\rm{and}}\,\,0\mu C\)
Explanation:
In steady state capacitor is fully charged and no current flows through it. \(\therefore \) No current passes through \(4\Omega \) \(\frac{1}{{{R_{eff}}}} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} = \frac{{6 + 3 + 2}}{6} = \frac{{11}}{6}\) \( \Rightarrow {R_{eff}} = \frac{6}{{11}}\Omega \) \({\rm{Current = }}\frac{{6 \times 11}}{6} = 11A\) \(Q = CV = 0.5 \times {10^{ - 6}} \times 6 = 3.0 \times {10^{ - 6}}C = 3\,\mu C\)
PHXII03:CURRENT ELECTRICITY
357492
Consider the circuit given below and calculate the charge on the capacitors.
1 \(5\,\mu C\)
2 \(1\,\mu C\)
3 \(3\,\mu C\)
4 \(7\,\mu C\)
Explanation:
\(I = \frac{3}{{6 \times {{10}^3}}} = 0.5 \times {10^{ - 3}}\;A\) \({V}_{{AD}}={IR}=0.5 \times 10^{-3} \times(2+1) \times 10^{3}=1.5 {~V}\) The equivalent capacitance of the given capacitors will be, \(\dfrac{1}{{C}_{{cq}}}=\dfrac{1}{{C}_{1}}+\dfrac{1}{{C}_{2}}=\dfrac{1}{1}+\dfrac{1}{2}=\dfrac{3}{2}\) \(\Rightarrow {C}_{{eq}}=\dfrac{2}{3} \mu {F}\) \(\therefore\) Charge on the capacitors, \({Q}=\dfrac{2 \times 10^{-6} \times 1.5}{3}=2 \times 10^{-6} \times 0.5=1 \mu {C}\)
PHXII03:CURRENT ELECTRICITY
357493
When the key \({K}\) is pressed at time \({t=0}\), which of the following is true about the current \({I}\) in the resistor \({A B}\) ?
1 2 \(mA\) all the time
2 oscillates between 1 \(mA\) and 2 \(mA\)
3 1 \(mA\) all the time
4 At \(t = 0\), \(I = 2\,\,mA\) and reduces to 1 \(mA\) finally
Explanation:
At \({t=0, I=\dfrac{2 {~V}}{1 {k} \Omega}=2 {~mA}}\) When capacitor is fully charged it will behave as open circuit, Net resistance of circuit, \({R=1 {k} \Omega+1 {k} \Omega=2 {k} \Omega}\) \(\therefore {\text{Current}} = \frac{{2\;V}}{{2\,k\,\Omega }} = 1\;\,mA\) Finally current will reduce to 1 mA . So correct option is (4)
PHXII03:CURRENT ELECTRICITY
357494
The circuit shown here has two batteries of 8.0\(V\) and 16.0 \(V\) and three resistors \(3\Omega \,,9\Omega \) and \(9\Omega \) and a capacitor \(5.0\,\mu F.\) How much is the current I in the circuit in steady state?
1 1.6 \(A\)
2 2.5 \(A\)
3 0.67 \(A\)
4 0.25 \(A\)
Explanation:
In the steady state current does not flow hrough the capacitor. The current in the outer circuit is \(16V - 9i - 3i - 8V = 0\) \(i = \frac{8}{{12}} = 0.67A\)
In steady state, current through \(AB = 0\) \( \Rightarrow {V_{AB}} = {V_{CD}}\) The current through outermost loop is \(i = \frac{E}{{r + {r_2}}}\) \( \Rightarrow {V_{AB}} = \left( {\frac{E}{{r + {r_2}}} \times {r_2}} \right)\) As potential difference across \({r_1}\) is zero. \( \Rightarrow Q = C{V_{AB}} = CE\left( {\frac{{{r_2}}}{{r + {r_2}}}} \right)\)
PHXII03:CURRENT ELECTRICITY
357491
In the given circuit diagram the current through the battery and the charge on the capacitor respectively in steady state is
1 \(11A\,\,{\rm{and}}\,\,{\rm{3}}\mu C\)
2 \(1A\,\,{\rm{and}}\,\,{\rm{3}}\mu C\)
3 \(6A\,\,{\rm{and}}\,\,0\mu C\)
4 \(17A\,\,{\rm{and}}\,\,0\mu C\)
Explanation:
In steady state capacitor is fully charged and no current flows through it. \(\therefore \) No current passes through \(4\Omega \) \(\frac{1}{{{R_{eff}}}} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} = \frac{{6 + 3 + 2}}{6} = \frac{{11}}{6}\) \( \Rightarrow {R_{eff}} = \frac{6}{{11}}\Omega \) \({\rm{Current = }}\frac{{6 \times 11}}{6} = 11A\) \(Q = CV = 0.5 \times {10^{ - 6}} \times 6 = 3.0 \times {10^{ - 6}}C = 3\,\mu C\)
PHXII03:CURRENT ELECTRICITY
357492
Consider the circuit given below and calculate the charge on the capacitors.
1 \(5\,\mu C\)
2 \(1\,\mu C\)
3 \(3\,\mu C\)
4 \(7\,\mu C\)
Explanation:
\(I = \frac{3}{{6 \times {{10}^3}}} = 0.5 \times {10^{ - 3}}\;A\) \({V}_{{AD}}={IR}=0.5 \times 10^{-3} \times(2+1) \times 10^{3}=1.5 {~V}\) The equivalent capacitance of the given capacitors will be, \(\dfrac{1}{{C}_{{cq}}}=\dfrac{1}{{C}_{1}}+\dfrac{1}{{C}_{2}}=\dfrac{1}{1}+\dfrac{1}{2}=\dfrac{3}{2}\) \(\Rightarrow {C}_{{eq}}=\dfrac{2}{3} \mu {F}\) \(\therefore\) Charge on the capacitors, \({Q}=\dfrac{2 \times 10^{-6} \times 1.5}{3}=2 \times 10^{-6} \times 0.5=1 \mu {C}\)
PHXII03:CURRENT ELECTRICITY
357493
When the key \({K}\) is pressed at time \({t=0}\), which of the following is true about the current \({I}\) in the resistor \({A B}\) ?
1 2 \(mA\) all the time
2 oscillates between 1 \(mA\) and 2 \(mA\)
3 1 \(mA\) all the time
4 At \(t = 0\), \(I = 2\,\,mA\) and reduces to 1 \(mA\) finally
Explanation:
At \({t=0, I=\dfrac{2 {~V}}{1 {k} \Omega}=2 {~mA}}\) When capacitor is fully charged it will behave as open circuit, Net resistance of circuit, \({R=1 {k} \Omega+1 {k} \Omega=2 {k} \Omega}\) \(\therefore {\text{Current}} = \frac{{2\;V}}{{2\,k\,\Omega }} = 1\;\,mA\) Finally current will reduce to 1 mA . So correct option is (4)
PHXII03:CURRENT ELECTRICITY
357494
The circuit shown here has two batteries of 8.0\(V\) and 16.0 \(V\) and three resistors \(3\Omega \,,9\Omega \) and \(9\Omega \) and a capacitor \(5.0\,\mu F.\) How much is the current I in the circuit in steady state?
1 1.6 \(A\)
2 2.5 \(A\)
3 0.67 \(A\)
4 0.25 \(A\)
Explanation:
In the steady state current does not flow hrough the capacitor. The current in the outer circuit is \(16V - 9i - 3i - 8V = 0\) \(i = \frac{8}{{12}} = 0.67A\)
In steady state, current through \(AB = 0\) \( \Rightarrow {V_{AB}} = {V_{CD}}\) The current through outermost loop is \(i = \frac{E}{{r + {r_2}}}\) \( \Rightarrow {V_{AB}} = \left( {\frac{E}{{r + {r_2}}} \times {r_2}} \right)\) As potential difference across \({r_1}\) is zero. \( \Rightarrow Q = C{V_{AB}} = CE\left( {\frac{{{r_2}}}{{r + {r_2}}}} \right)\)
PHXII03:CURRENT ELECTRICITY
357491
In the given circuit diagram the current through the battery and the charge on the capacitor respectively in steady state is
1 \(11A\,\,{\rm{and}}\,\,{\rm{3}}\mu C\)
2 \(1A\,\,{\rm{and}}\,\,{\rm{3}}\mu C\)
3 \(6A\,\,{\rm{and}}\,\,0\mu C\)
4 \(17A\,\,{\rm{and}}\,\,0\mu C\)
Explanation:
In steady state capacitor is fully charged and no current flows through it. \(\therefore \) No current passes through \(4\Omega \) \(\frac{1}{{{R_{eff}}}} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} = \frac{{6 + 3 + 2}}{6} = \frac{{11}}{6}\) \( \Rightarrow {R_{eff}} = \frac{6}{{11}}\Omega \) \({\rm{Current = }}\frac{{6 \times 11}}{6} = 11A\) \(Q = CV = 0.5 \times {10^{ - 6}} \times 6 = 3.0 \times {10^{ - 6}}C = 3\,\mu C\)
PHXII03:CURRENT ELECTRICITY
357492
Consider the circuit given below and calculate the charge on the capacitors.
1 \(5\,\mu C\)
2 \(1\,\mu C\)
3 \(3\,\mu C\)
4 \(7\,\mu C\)
Explanation:
\(I = \frac{3}{{6 \times {{10}^3}}} = 0.5 \times {10^{ - 3}}\;A\) \({V}_{{AD}}={IR}=0.5 \times 10^{-3} \times(2+1) \times 10^{3}=1.5 {~V}\) The equivalent capacitance of the given capacitors will be, \(\dfrac{1}{{C}_{{cq}}}=\dfrac{1}{{C}_{1}}+\dfrac{1}{{C}_{2}}=\dfrac{1}{1}+\dfrac{1}{2}=\dfrac{3}{2}\) \(\Rightarrow {C}_{{eq}}=\dfrac{2}{3} \mu {F}\) \(\therefore\) Charge on the capacitors, \({Q}=\dfrac{2 \times 10^{-6} \times 1.5}{3}=2 \times 10^{-6} \times 0.5=1 \mu {C}\)
PHXII03:CURRENT ELECTRICITY
357493
When the key \({K}\) is pressed at time \({t=0}\), which of the following is true about the current \({I}\) in the resistor \({A B}\) ?
1 2 \(mA\) all the time
2 oscillates between 1 \(mA\) and 2 \(mA\)
3 1 \(mA\) all the time
4 At \(t = 0\), \(I = 2\,\,mA\) and reduces to 1 \(mA\) finally
Explanation:
At \({t=0, I=\dfrac{2 {~V}}{1 {k} \Omega}=2 {~mA}}\) When capacitor is fully charged it will behave as open circuit, Net resistance of circuit, \({R=1 {k} \Omega+1 {k} \Omega=2 {k} \Omega}\) \(\therefore {\text{Current}} = \frac{{2\;V}}{{2\,k\,\Omega }} = 1\;\,mA\) Finally current will reduce to 1 mA . So correct option is (4)
PHXII03:CURRENT ELECTRICITY
357494
The circuit shown here has two batteries of 8.0\(V\) and 16.0 \(V\) and three resistors \(3\Omega \,,9\Omega \) and \(9\Omega \) and a capacitor \(5.0\,\mu F.\) How much is the current I in the circuit in steady state?
1 1.6 \(A\)
2 2.5 \(A\)
3 0.67 \(A\)
4 0.25 \(A\)
Explanation:
In the steady state current does not flow hrough the capacitor. The current in the outer circuit is \(16V - 9i - 3i - 8V = 0\) \(i = \frac{8}{{12}} = 0.67A\)
