NEET Test Series from KOTA - 10 Papers In MS WORD
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Current Electricity
268498
If in the circuit shown below, the internalresistnce of the battery is \(1 \Omega\) and \(V_{p}\) and \(V_{o}\) are the potentials at \(P\) and \(Q\) respectively, the potential difference between the points \(P\) and \(Q\) is
1 \(9 \mathrm{~V}\)
2 \(11 \mathrm{~V}\)
3 \(7 \mathrm{~V}\)
4 \(6 \mathrm{~V}\)
Explanation:
\(\mathrm{i}=\frac{V}{R}\)
Current Electricity
268499
Voltmeter reading in the given circuit is\({ }^{142}\) (voltmeter is ideal)
1 \(6 V \)
2 \(8 V\)
3 \(10 V \)
4 \(14 V\)
Explanation:
\(V=\mathrm{iR}\)
Current Electricity
268500
For a cell the graph between thep.d(v) across the terminals of the cells and the current (I) drawn from the cell as shown. The emf and inernal resistance is
1 \(\frac{3}{2} \Omega\)
2 \(\frac{1}{3} \Omega\)
3 \(3 \Omega\)
4 \(\frac{2}{3} \Omega\)
Explanation:
\(\mathrm{i}=\frac{E}{r}\)
Current Electricity
268501
Theminimum number of cells in mixed grouping required to produced a maximum current of 1 A through external resistance of \(20 \Omega\) given the emf of each cell is \(2 \mathrm{~V}\) and internal resistance \(1 \Omega\) is
1 25
2 20
3 16
4 30
Explanation:
No. of cells\(=m \times n\) \(\mathrm{i}_{\max }=\mathrm{mR}=\mathrm{nr}\)
268498
If in the circuit shown below, the internalresistnce of the battery is \(1 \Omega\) and \(V_{p}\) and \(V_{o}\) are the potentials at \(P\) and \(Q\) respectively, the potential difference between the points \(P\) and \(Q\) is
1 \(9 \mathrm{~V}\)
2 \(11 \mathrm{~V}\)
3 \(7 \mathrm{~V}\)
4 \(6 \mathrm{~V}\)
Explanation:
\(\mathrm{i}=\frac{V}{R}\)
Current Electricity
268499
Voltmeter reading in the given circuit is\({ }^{142}\) (voltmeter is ideal)
1 \(6 V \)
2 \(8 V\)
3 \(10 V \)
4 \(14 V\)
Explanation:
\(V=\mathrm{iR}\)
Current Electricity
268500
For a cell the graph between thep.d(v) across the terminals of the cells and the current (I) drawn from the cell as shown. The emf and inernal resistance is
1 \(\frac{3}{2} \Omega\)
2 \(\frac{1}{3} \Omega\)
3 \(3 \Omega\)
4 \(\frac{2}{3} \Omega\)
Explanation:
\(\mathrm{i}=\frac{E}{r}\)
Current Electricity
268501
Theminimum number of cells in mixed grouping required to produced a maximum current of 1 A through external resistance of \(20 \Omega\) given the emf of each cell is \(2 \mathrm{~V}\) and internal resistance \(1 \Omega\) is
1 25
2 20
3 16
4 30
Explanation:
No. of cells\(=m \times n\) \(\mathrm{i}_{\max }=\mathrm{mR}=\mathrm{nr}\)
268498
If in the circuit shown below, the internalresistnce of the battery is \(1 \Omega\) and \(V_{p}\) and \(V_{o}\) are the potentials at \(P\) and \(Q\) respectively, the potential difference between the points \(P\) and \(Q\) is
1 \(9 \mathrm{~V}\)
2 \(11 \mathrm{~V}\)
3 \(7 \mathrm{~V}\)
4 \(6 \mathrm{~V}\)
Explanation:
\(\mathrm{i}=\frac{V}{R}\)
Current Electricity
268499
Voltmeter reading in the given circuit is\({ }^{142}\) (voltmeter is ideal)
1 \(6 V \)
2 \(8 V\)
3 \(10 V \)
4 \(14 V\)
Explanation:
\(V=\mathrm{iR}\)
Current Electricity
268500
For a cell the graph between thep.d(v) across the terminals of the cells and the current (I) drawn from the cell as shown. The emf and inernal resistance is
1 \(\frac{3}{2} \Omega\)
2 \(\frac{1}{3} \Omega\)
3 \(3 \Omega\)
4 \(\frac{2}{3} \Omega\)
Explanation:
\(\mathrm{i}=\frac{E}{r}\)
Current Electricity
268501
Theminimum number of cells in mixed grouping required to produced a maximum current of 1 A through external resistance of \(20 \Omega\) given the emf of each cell is \(2 \mathrm{~V}\) and internal resistance \(1 \Omega\) is
1 25
2 20
3 16
4 30
Explanation:
No. of cells\(=m \times n\) \(\mathrm{i}_{\max }=\mathrm{mR}=\mathrm{nr}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Current Electricity
268498
If in the circuit shown below, the internalresistnce of the battery is \(1 \Omega\) and \(V_{p}\) and \(V_{o}\) are the potentials at \(P\) and \(Q\) respectively, the potential difference between the points \(P\) and \(Q\) is
1 \(9 \mathrm{~V}\)
2 \(11 \mathrm{~V}\)
3 \(7 \mathrm{~V}\)
4 \(6 \mathrm{~V}\)
Explanation:
\(\mathrm{i}=\frac{V}{R}\)
Current Electricity
268499
Voltmeter reading in the given circuit is\({ }^{142}\) (voltmeter is ideal)
1 \(6 V \)
2 \(8 V\)
3 \(10 V \)
4 \(14 V\)
Explanation:
\(V=\mathrm{iR}\)
Current Electricity
268500
For a cell the graph between thep.d(v) across the terminals of the cells and the current (I) drawn from the cell as shown. The emf and inernal resistance is
1 \(\frac{3}{2} \Omega\)
2 \(\frac{1}{3} \Omega\)
3 \(3 \Omega\)
4 \(\frac{2}{3} \Omega\)
Explanation:
\(\mathrm{i}=\frac{E}{r}\)
Current Electricity
268501
Theminimum number of cells in mixed grouping required to produced a maximum current of 1 A through external resistance of \(20 \Omega\) given the emf of each cell is \(2 \mathrm{~V}\) and internal resistance \(1 \Omega\) is
1 25
2 20
3 16
4 30
Explanation:
No. of cells\(=m \times n\) \(\mathrm{i}_{\max }=\mathrm{mR}=\mathrm{nr}\)