357234
A 9 \(V\) battery with internal resistance of \(0.5\Omega \) is connected across an infinite network as shown in the figure. All ammeters \({A_1},{A_2},{A_3}\) and voltmeter \(V\) are ideal. Choose correct statement
1 Reading of \({A_1}\) is 18 \(A\)
2 Reading of \(V\) is 9 \(V\)
3 Reading of \(V\) is 7 \(V\)
4 Reading of \({A_1}\) is 2 \(A\)
Explanation:
Let \(x\) be the equivalent resistance across \(PQ\) and hence the equivalent resistance across \(ST\) is also \(x\) . The given circuit can be redrawn as \( \Rightarrow {R_{PQ}} = 1 + \frac{{4x}}{{4 + x}} + 1\) As \({{\text{R}}_{{\text{PQ}}}} = x = 2 + \frac{{4x}}{{4 + x}}\) \({x^2} - 2x - 8 = 0\quad \Rightarrow x = 4\Omega \) The current through the battery is \(i = \frac{{9V}}{{0.5 + 4}} = 2A\) The reading of \({A_1} = 2A\) The reading of \(V = 9V - 2\left( {0.5} \right) = 8V\)
JEE - 2017
PHXII03:CURRENT ELECTRICITY
357235
In the following circuit, the resistance of a voltmeter is \(10,000\,\Omega \) and that of an ammeter is \(20\,\Omega \). If the reading of an ammeter is 0.1 amp. and that of voltmeter is 12 volt, then the value of \(R\) is:
357236
The reading in the ideal voltmeter \((V)\) shown in the given circuit diagram is
1 \(10\,V\)
2 \(0\,V\)
3 \(5\,V\)
4 \(3\,V\)
Explanation:
Here 8 cells are in series, So, \({i_{{\text{net }}}} = \frac{{{E_{{\text{eff }}}}}}{{{R_{{\text{eff }}}}}} = \frac{{nE}}{{nr}} = \frac{{8 \times 5}}{{8 \times 0.2}} = \frac{5}{{0.2}} = 25\;A\) Reading of voltmeter, \(V=E-i r\) \( = 5 - 0.2 \times 25 = 0\;V\)
JEE - 2024
PHXII03:CURRENT ELECTRICITY
357237
In the given circuit, the voltmeter and the electric cell are ideal. The reading of the voltmeter is
1 \(4\,V\)
2 \(1\,V\)
3 \(6\,V\)
4 \(8\,V\)
Explanation:
The electric current through ideal voltmeter is zero. According to loop rule, \({E-1 \times I-1 \times I=0 \Rightarrow I=\dfrac{E}{2}=\dfrac{2}{2}=1 {~A}}\) Reading of the voltmeter \({=V_{A}-V_{B}=[1 \times I]=[1 \times 1]=1 {~V}}\)
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PHXII03:CURRENT ELECTRICITY
357234
A 9 \(V\) battery with internal resistance of \(0.5\Omega \) is connected across an infinite network as shown in the figure. All ammeters \({A_1},{A_2},{A_3}\) and voltmeter \(V\) are ideal. Choose correct statement
1 Reading of \({A_1}\) is 18 \(A\)
2 Reading of \(V\) is 9 \(V\)
3 Reading of \(V\) is 7 \(V\)
4 Reading of \({A_1}\) is 2 \(A\)
Explanation:
Let \(x\) be the equivalent resistance across \(PQ\) and hence the equivalent resistance across \(ST\) is also \(x\) . The given circuit can be redrawn as \( \Rightarrow {R_{PQ}} = 1 + \frac{{4x}}{{4 + x}} + 1\) As \({{\text{R}}_{{\text{PQ}}}} = x = 2 + \frac{{4x}}{{4 + x}}\) \({x^2} - 2x - 8 = 0\quad \Rightarrow x = 4\Omega \) The current through the battery is \(i = \frac{{9V}}{{0.5 + 4}} = 2A\) The reading of \({A_1} = 2A\) The reading of \(V = 9V - 2\left( {0.5} \right) = 8V\)
JEE - 2017
PHXII03:CURRENT ELECTRICITY
357235
In the following circuit, the resistance of a voltmeter is \(10,000\,\Omega \) and that of an ammeter is \(20\,\Omega \). If the reading of an ammeter is 0.1 amp. and that of voltmeter is 12 volt, then the value of \(R\) is:
357236
The reading in the ideal voltmeter \((V)\) shown in the given circuit diagram is
1 \(10\,V\)
2 \(0\,V\)
3 \(5\,V\)
4 \(3\,V\)
Explanation:
Here 8 cells are in series, So, \({i_{{\text{net }}}} = \frac{{{E_{{\text{eff }}}}}}{{{R_{{\text{eff }}}}}} = \frac{{nE}}{{nr}} = \frac{{8 \times 5}}{{8 \times 0.2}} = \frac{5}{{0.2}} = 25\;A\) Reading of voltmeter, \(V=E-i r\) \( = 5 - 0.2 \times 25 = 0\;V\)
JEE - 2024
PHXII03:CURRENT ELECTRICITY
357237
In the given circuit, the voltmeter and the electric cell are ideal. The reading of the voltmeter is
1 \(4\,V\)
2 \(1\,V\)
3 \(6\,V\)
4 \(8\,V\)
Explanation:
The electric current through ideal voltmeter is zero. According to loop rule, \({E-1 \times I-1 \times I=0 \Rightarrow I=\dfrac{E}{2}=\dfrac{2}{2}=1 {~A}}\) Reading of the voltmeter \({=V_{A}-V_{B}=[1 \times I]=[1 \times 1]=1 {~V}}\)
357234
A 9 \(V\) battery with internal resistance of \(0.5\Omega \) is connected across an infinite network as shown in the figure. All ammeters \({A_1},{A_2},{A_3}\) and voltmeter \(V\) are ideal. Choose correct statement
1 Reading of \({A_1}\) is 18 \(A\)
2 Reading of \(V\) is 9 \(V\)
3 Reading of \(V\) is 7 \(V\)
4 Reading of \({A_1}\) is 2 \(A\)
Explanation:
Let \(x\) be the equivalent resistance across \(PQ\) and hence the equivalent resistance across \(ST\) is also \(x\) . The given circuit can be redrawn as \( \Rightarrow {R_{PQ}} = 1 + \frac{{4x}}{{4 + x}} + 1\) As \({{\text{R}}_{{\text{PQ}}}} = x = 2 + \frac{{4x}}{{4 + x}}\) \({x^2} - 2x - 8 = 0\quad \Rightarrow x = 4\Omega \) The current through the battery is \(i = \frac{{9V}}{{0.5 + 4}} = 2A\) The reading of \({A_1} = 2A\) The reading of \(V = 9V - 2\left( {0.5} \right) = 8V\)
JEE - 2017
PHXII03:CURRENT ELECTRICITY
357235
In the following circuit, the resistance of a voltmeter is \(10,000\,\Omega \) and that of an ammeter is \(20\,\Omega \). If the reading of an ammeter is 0.1 amp. and that of voltmeter is 12 volt, then the value of \(R\) is:
357236
The reading in the ideal voltmeter \((V)\) shown in the given circuit diagram is
1 \(10\,V\)
2 \(0\,V\)
3 \(5\,V\)
4 \(3\,V\)
Explanation:
Here 8 cells are in series, So, \({i_{{\text{net }}}} = \frac{{{E_{{\text{eff }}}}}}{{{R_{{\text{eff }}}}}} = \frac{{nE}}{{nr}} = \frac{{8 \times 5}}{{8 \times 0.2}} = \frac{5}{{0.2}} = 25\;A\) Reading of voltmeter, \(V=E-i r\) \( = 5 - 0.2 \times 25 = 0\;V\)
JEE - 2024
PHXII03:CURRENT ELECTRICITY
357237
In the given circuit, the voltmeter and the electric cell are ideal. The reading of the voltmeter is
1 \(4\,V\)
2 \(1\,V\)
3 \(6\,V\)
4 \(8\,V\)
Explanation:
The electric current through ideal voltmeter is zero. According to loop rule, \({E-1 \times I-1 \times I=0 \Rightarrow I=\dfrac{E}{2}=\dfrac{2}{2}=1 {~A}}\) Reading of the voltmeter \({=V_{A}-V_{B}=[1 \times I]=[1 \times 1]=1 {~V}}\)
357234
A 9 \(V\) battery with internal resistance of \(0.5\Omega \) is connected across an infinite network as shown in the figure. All ammeters \({A_1},{A_2},{A_3}\) and voltmeter \(V\) are ideal. Choose correct statement
1 Reading of \({A_1}\) is 18 \(A\)
2 Reading of \(V\) is 9 \(V\)
3 Reading of \(V\) is 7 \(V\)
4 Reading of \({A_1}\) is 2 \(A\)
Explanation:
Let \(x\) be the equivalent resistance across \(PQ\) and hence the equivalent resistance across \(ST\) is also \(x\) . The given circuit can be redrawn as \( \Rightarrow {R_{PQ}} = 1 + \frac{{4x}}{{4 + x}} + 1\) As \({{\text{R}}_{{\text{PQ}}}} = x = 2 + \frac{{4x}}{{4 + x}}\) \({x^2} - 2x - 8 = 0\quad \Rightarrow x = 4\Omega \) The current through the battery is \(i = \frac{{9V}}{{0.5 + 4}} = 2A\) The reading of \({A_1} = 2A\) The reading of \(V = 9V - 2\left( {0.5} \right) = 8V\)
JEE - 2017
PHXII03:CURRENT ELECTRICITY
357235
In the following circuit, the resistance of a voltmeter is \(10,000\,\Omega \) and that of an ammeter is \(20\,\Omega \). If the reading of an ammeter is 0.1 amp. and that of voltmeter is 12 volt, then the value of \(R\) is:
357236
The reading in the ideal voltmeter \((V)\) shown in the given circuit diagram is
1 \(10\,V\)
2 \(0\,V\)
3 \(5\,V\)
4 \(3\,V\)
Explanation:
Here 8 cells are in series, So, \({i_{{\text{net }}}} = \frac{{{E_{{\text{eff }}}}}}{{{R_{{\text{eff }}}}}} = \frac{{nE}}{{nr}} = \frac{{8 \times 5}}{{8 \times 0.2}} = \frac{5}{{0.2}} = 25\;A\) Reading of voltmeter, \(V=E-i r\) \( = 5 - 0.2 \times 25 = 0\;V\)
JEE - 2024
PHXII03:CURRENT ELECTRICITY
357237
In the given circuit, the voltmeter and the electric cell are ideal. The reading of the voltmeter is
1 \(4\,V\)
2 \(1\,V\)
3 \(6\,V\)
4 \(8\,V\)
Explanation:
The electric current through ideal voltmeter is zero. According to loop rule, \({E-1 \times I-1 \times I=0 \Rightarrow I=\dfrac{E}{2}=\dfrac{2}{2}=1 {~A}}\) Reading of the voltmeter \({=V_{A}-V_{B}=[1 \times I]=[1 \times 1]=1 {~V}}\)
