356862
For the ladder given find \({\varepsilon _{eq}}\) and \({r_{eq}}\) across \(AB\).
1 \(4V,\,2\Omega \)
2 \(2V,\,2\Omega \)
3 \(4V,\,1\Omega \)
4 \(2V,\,6\Omega \)
Explanation:
Let \(\varepsilon \) and \(r\) be the equivalent emf and equivalent resistance across \(AB\). We can consider the circuit as Across \(CD\) also equivalent emf and resistance are \(\varepsilon \) and \(r\). \(\varepsilon ' = \frac{{\frac{\varepsilon }{r} + \frac{0}{2}}}{{\frac{1}{r} + \frac{1}{2}}} = \frac{{2\varepsilon }}{{2 + r}}\) \(\frac{1}{{r'}} = \frac{1}{r} + \frac{1}{2} \Rightarrow r' = \frac{{2r}}{{2 + r}}\) we can draw the circuit as we have taken that \(\varepsilon ' + 2 = \varepsilon \,\,{\rm{and}}\,\,r' + 1 = r\) \( \Rightarrow \frac{{2\varepsilon }}{{2 + r}} + 2 = \varepsilon \Rightarrow \varepsilon r - 4 - 2r = 0\,\,\,\,\,\left( 1 \right)\) \(r' + 1 = r \Rightarrow {r^2} - r - 2 = 0 \to \left( 2 \right)\) From (1) & (2) \(r = 2\Omega ,\quad \varepsilon = 4V\)
PHXII03:CURRENT ELECTRICITY
356863
Two batteries with e.m.f 12\(V\) and 13\(V\) are connected in parallel across a load resistor of \(10\Omega \). The internal resistances of the two batteries are \(1\Omega \,\) and \(2\Omega \) respectively. The voltage across the load lies between.
1 11.5 \(V\) and 11.6 \(V\)
2 11.4 \(V\) and 11.5 \(V\)
3 11.7 \(V\) and 11.8 \(V\)
4 11.6 \(V\) and 11.7 \(V\)
Explanation:
The equivalent emf and internal resistance are \({\varepsilon _{eq}} = \frac{{\frac{{{\varepsilon _1}}}{{{r_1}}} + \frac{{{\varepsilon _2}}}{{{r_2}}}}}{{\frac{1}{{{r_1}}} + \frac{1}{{{r_2}}}}}\) \(\frac{{\frac{{12}}{1} + \frac{{13}}{2}}}{{\frac{1}{1} + \frac{1}{2}}} = \frac{{37}}{3}V\) \({r_{eq}} = \frac{{2 \times 1}}{{2 + 1}} = \frac{2}{3}\Omega \) The equivalent circuit is \(i = \frac{{37/3}}{{10 + \frac{2}{3}}} = \frac{{37}}{{32}}\) \(\therefore {V_{10\Omega }} = i \times 10 = \frac{{37}}{{32}} \times 10 = 11.56\) volt
PHXII03:CURRENT ELECTRICITY
356864
Assertion : In order to have maximum current a series combinations of cells is used where their internal resistance is much smaller than the external resistance. Reason : Current flowing through the circuit is inversely proportional to total resistance.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
\(I=(m E / E+n r)\) In series connection, total emf of cells adds, and their internal resistances sum. When cell internal resistance \((n r)\) is much smaller than external resistance \((R)\), the current formula simplifies to approximately \(n E / R\), maximizing current with minimal impact from internal resistance. So, series connection gives maximum current. So correct option is (1).
PHXII03:CURRENT ELECTRICITY
356865
Four identical cells of emf \(E\) and internal resistance \(r\) are to be connected in series. Suppose if one of the cell is connected wrongly, the equivalent emf and effective internal resistance of the combination is
1 \(4E\,{\rm{and}}\,2r\)
2 \(2E\,{\rm{and}}\,2r\)
3 \(4E\,{\rm{and}}\,4r\)
4 \(2E\,{\rm{and}}\,4r\)
Explanation:
When 3 cells are connected correctly and one incorrectly, then equivalent emf of the combination is \({E_{eq}} = (E + E + E - E) = 2E\) and effective internal resistance of the combination is \({r_{eff}} = (r + r + r + r) = 4r\)
KCET - 2015
PHXII03:CURRENT ELECTRICITY
356866
In the electric circuit shown, each cell has an emf of \(2\,\;V\) and internal resistance is \(1 \Omega\). The external resistance is \(2\,\Omega \). The value of the current \(I\) is (in ampere)
356862
For the ladder given find \({\varepsilon _{eq}}\) and \({r_{eq}}\) across \(AB\).
1 \(4V,\,2\Omega \)
2 \(2V,\,2\Omega \)
3 \(4V,\,1\Omega \)
4 \(2V,\,6\Omega \)
Explanation:
Let \(\varepsilon \) and \(r\) be the equivalent emf and equivalent resistance across \(AB\). We can consider the circuit as Across \(CD\) also equivalent emf and resistance are \(\varepsilon \) and \(r\). \(\varepsilon ' = \frac{{\frac{\varepsilon }{r} + \frac{0}{2}}}{{\frac{1}{r} + \frac{1}{2}}} = \frac{{2\varepsilon }}{{2 + r}}\) \(\frac{1}{{r'}} = \frac{1}{r} + \frac{1}{2} \Rightarrow r' = \frac{{2r}}{{2 + r}}\) we can draw the circuit as we have taken that \(\varepsilon ' + 2 = \varepsilon \,\,{\rm{and}}\,\,r' + 1 = r\) \( \Rightarrow \frac{{2\varepsilon }}{{2 + r}} + 2 = \varepsilon \Rightarrow \varepsilon r - 4 - 2r = 0\,\,\,\,\,\left( 1 \right)\) \(r' + 1 = r \Rightarrow {r^2} - r - 2 = 0 \to \left( 2 \right)\) From (1) & (2) \(r = 2\Omega ,\quad \varepsilon = 4V\)
PHXII03:CURRENT ELECTRICITY
356863
Two batteries with e.m.f 12\(V\) and 13\(V\) are connected in parallel across a load resistor of \(10\Omega \). The internal resistances of the two batteries are \(1\Omega \,\) and \(2\Omega \) respectively. The voltage across the load lies between.
1 11.5 \(V\) and 11.6 \(V\)
2 11.4 \(V\) and 11.5 \(V\)
3 11.7 \(V\) and 11.8 \(V\)
4 11.6 \(V\) and 11.7 \(V\)
Explanation:
The equivalent emf and internal resistance are \({\varepsilon _{eq}} = \frac{{\frac{{{\varepsilon _1}}}{{{r_1}}} + \frac{{{\varepsilon _2}}}{{{r_2}}}}}{{\frac{1}{{{r_1}}} + \frac{1}{{{r_2}}}}}\) \(\frac{{\frac{{12}}{1} + \frac{{13}}{2}}}{{\frac{1}{1} + \frac{1}{2}}} = \frac{{37}}{3}V\) \({r_{eq}} = \frac{{2 \times 1}}{{2 + 1}} = \frac{2}{3}\Omega \) The equivalent circuit is \(i = \frac{{37/3}}{{10 + \frac{2}{3}}} = \frac{{37}}{{32}}\) \(\therefore {V_{10\Omega }} = i \times 10 = \frac{{37}}{{32}} \times 10 = 11.56\) volt
PHXII03:CURRENT ELECTRICITY
356864
Assertion : In order to have maximum current a series combinations of cells is used where their internal resistance is much smaller than the external resistance. Reason : Current flowing through the circuit is inversely proportional to total resistance.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
\(I=(m E / E+n r)\) In series connection, total emf of cells adds, and their internal resistances sum. When cell internal resistance \((n r)\) is much smaller than external resistance \((R)\), the current formula simplifies to approximately \(n E / R\), maximizing current with minimal impact from internal resistance. So, series connection gives maximum current. So correct option is (1).
PHXII03:CURRENT ELECTRICITY
356865
Four identical cells of emf \(E\) and internal resistance \(r\) are to be connected in series. Suppose if one of the cell is connected wrongly, the equivalent emf and effective internal resistance of the combination is
1 \(4E\,{\rm{and}}\,2r\)
2 \(2E\,{\rm{and}}\,2r\)
3 \(4E\,{\rm{and}}\,4r\)
4 \(2E\,{\rm{and}}\,4r\)
Explanation:
When 3 cells are connected correctly and one incorrectly, then equivalent emf of the combination is \({E_{eq}} = (E + E + E - E) = 2E\) and effective internal resistance of the combination is \({r_{eff}} = (r + r + r + r) = 4r\)
KCET - 2015
PHXII03:CURRENT ELECTRICITY
356866
In the electric circuit shown, each cell has an emf of \(2\,\;V\) and internal resistance is \(1 \Omega\). The external resistance is \(2\,\Omega \). The value of the current \(I\) is (in ampere)
356862
For the ladder given find \({\varepsilon _{eq}}\) and \({r_{eq}}\) across \(AB\).
1 \(4V,\,2\Omega \)
2 \(2V,\,2\Omega \)
3 \(4V,\,1\Omega \)
4 \(2V,\,6\Omega \)
Explanation:
Let \(\varepsilon \) and \(r\) be the equivalent emf and equivalent resistance across \(AB\). We can consider the circuit as Across \(CD\) also equivalent emf and resistance are \(\varepsilon \) and \(r\). \(\varepsilon ' = \frac{{\frac{\varepsilon }{r} + \frac{0}{2}}}{{\frac{1}{r} + \frac{1}{2}}} = \frac{{2\varepsilon }}{{2 + r}}\) \(\frac{1}{{r'}} = \frac{1}{r} + \frac{1}{2} \Rightarrow r' = \frac{{2r}}{{2 + r}}\) we can draw the circuit as we have taken that \(\varepsilon ' + 2 = \varepsilon \,\,{\rm{and}}\,\,r' + 1 = r\) \( \Rightarrow \frac{{2\varepsilon }}{{2 + r}} + 2 = \varepsilon \Rightarrow \varepsilon r - 4 - 2r = 0\,\,\,\,\,\left( 1 \right)\) \(r' + 1 = r \Rightarrow {r^2} - r - 2 = 0 \to \left( 2 \right)\) From (1) & (2) \(r = 2\Omega ,\quad \varepsilon = 4V\)
PHXII03:CURRENT ELECTRICITY
356863
Two batteries with e.m.f 12\(V\) and 13\(V\) are connected in parallel across a load resistor of \(10\Omega \). The internal resistances of the two batteries are \(1\Omega \,\) and \(2\Omega \) respectively. The voltage across the load lies between.
1 11.5 \(V\) and 11.6 \(V\)
2 11.4 \(V\) and 11.5 \(V\)
3 11.7 \(V\) and 11.8 \(V\)
4 11.6 \(V\) and 11.7 \(V\)
Explanation:
The equivalent emf and internal resistance are \({\varepsilon _{eq}} = \frac{{\frac{{{\varepsilon _1}}}{{{r_1}}} + \frac{{{\varepsilon _2}}}{{{r_2}}}}}{{\frac{1}{{{r_1}}} + \frac{1}{{{r_2}}}}}\) \(\frac{{\frac{{12}}{1} + \frac{{13}}{2}}}{{\frac{1}{1} + \frac{1}{2}}} = \frac{{37}}{3}V\) \({r_{eq}} = \frac{{2 \times 1}}{{2 + 1}} = \frac{2}{3}\Omega \) The equivalent circuit is \(i = \frac{{37/3}}{{10 + \frac{2}{3}}} = \frac{{37}}{{32}}\) \(\therefore {V_{10\Omega }} = i \times 10 = \frac{{37}}{{32}} \times 10 = 11.56\) volt
PHXII03:CURRENT ELECTRICITY
356864
Assertion : In order to have maximum current a series combinations of cells is used where their internal resistance is much smaller than the external resistance. Reason : Current flowing through the circuit is inversely proportional to total resistance.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
\(I=(m E / E+n r)\) In series connection, total emf of cells adds, and their internal resistances sum. When cell internal resistance \((n r)\) is much smaller than external resistance \((R)\), the current formula simplifies to approximately \(n E / R\), maximizing current with minimal impact from internal resistance. So, series connection gives maximum current. So correct option is (1).
PHXII03:CURRENT ELECTRICITY
356865
Four identical cells of emf \(E\) and internal resistance \(r\) are to be connected in series. Suppose if one of the cell is connected wrongly, the equivalent emf and effective internal resistance of the combination is
1 \(4E\,{\rm{and}}\,2r\)
2 \(2E\,{\rm{and}}\,2r\)
3 \(4E\,{\rm{and}}\,4r\)
4 \(2E\,{\rm{and}}\,4r\)
Explanation:
When 3 cells are connected correctly and one incorrectly, then equivalent emf of the combination is \({E_{eq}} = (E + E + E - E) = 2E\) and effective internal resistance of the combination is \({r_{eff}} = (r + r + r + r) = 4r\)
KCET - 2015
PHXII03:CURRENT ELECTRICITY
356866
In the electric circuit shown, each cell has an emf of \(2\,\;V\) and internal resistance is \(1 \Omega\). The external resistance is \(2\,\Omega \). The value of the current \(I\) is (in ampere)
NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXII03:CURRENT ELECTRICITY
356862
For the ladder given find \({\varepsilon _{eq}}\) and \({r_{eq}}\) across \(AB\).
1 \(4V,\,2\Omega \)
2 \(2V,\,2\Omega \)
3 \(4V,\,1\Omega \)
4 \(2V,\,6\Omega \)
Explanation:
Let \(\varepsilon \) and \(r\) be the equivalent emf and equivalent resistance across \(AB\). We can consider the circuit as Across \(CD\) also equivalent emf and resistance are \(\varepsilon \) and \(r\). \(\varepsilon ' = \frac{{\frac{\varepsilon }{r} + \frac{0}{2}}}{{\frac{1}{r} + \frac{1}{2}}} = \frac{{2\varepsilon }}{{2 + r}}\) \(\frac{1}{{r'}} = \frac{1}{r} + \frac{1}{2} \Rightarrow r' = \frac{{2r}}{{2 + r}}\) we can draw the circuit as we have taken that \(\varepsilon ' + 2 = \varepsilon \,\,{\rm{and}}\,\,r' + 1 = r\) \( \Rightarrow \frac{{2\varepsilon }}{{2 + r}} + 2 = \varepsilon \Rightarrow \varepsilon r - 4 - 2r = 0\,\,\,\,\,\left( 1 \right)\) \(r' + 1 = r \Rightarrow {r^2} - r - 2 = 0 \to \left( 2 \right)\) From (1) & (2) \(r = 2\Omega ,\quad \varepsilon = 4V\)
PHXII03:CURRENT ELECTRICITY
356863
Two batteries with e.m.f 12\(V\) and 13\(V\) are connected in parallel across a load resistor of \(10\Omega \). The internal resistances of the two batteries are \(1\Omega \,\) and \(2\Omega \) respectively. The voltage across the load lies between.
1 11.5 \(V\) and 11.6 \(V\)
2 11.4 \(V\) and 11.5 \(V\)
3 11.7 \(V\) and 11.8 \(V\)
4 11.6 \(V\) and 11.7 \(V\)
Explanation:
The equivalent emf and internal resistance are \({\varepsilon _{eq}} = \frac{{\frac{{{\varepsilon _1}}}{{{r_1}}} + \frac{{{\varepsilon _2}}}{{{r_2}}}}}{{\frac{1}{{{r_1}}} + \frac{1}{{{r_2}}}}}\) \(\frac{{\frac{{12}}{1} + \frac{{13}}{2}}}{{\frac{1}{1} + \frac{1}{2}}} = \frac{{37}}{3}V\) \({r_{eq}} = \frac{{2 \times 1}}{{2 + 1}} = \frac{2}{3}\Omega \) The equivalent circuit is \(i = \frac{{37/3}}{{10 + \frac{2}{3}}} = \frac{{37}}{{32}}\) \(\therefore {V_{10\Omega }} = i \times 10 = \frac{{37}}{{32}} \times 10 = 11.56\) volt
PHXII03:CURRENT ELECTRICITY
356864
Assertion : In order to have maximum current a series combinations of cells is used where their internal resistance is much smaller than the external resistance. Reason : Current flowing through the circuit is inversely proportional to total resistance.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
\(I=(m E / E+n r)\) In series connection, total emf of cells adds, and their internal resistances sum. When cell internal resistance \((n r)\) is much smaller than external resistance \((R)\), the current formula simplifies to approximately \(n E / R\), maximizing current with minimal impact from internal resistance. So, series connection gives maximum current. So correct option is (1).
PHXII03:CURRENT ELECTRICITY
356865
Four identical cells of emf \(E\) and internal resistance \(r\) are to be connected in series. Suppose if one of the cell is connected wrongly, the equivalent emf and effective internal resistance of the combination is
1 \(4E\,{\rm{and}}\,2r\)
2 \(2E\,{\rm{and}}\,2r\)
3 \(4E\,{\rm{and}}\,4r\)
4 \(2E\,{\rm{and}}\,4r\)
Explanation:
When 3 cells are connected correctly and one incorrectly, then equivalent emf of the combination is \({E_{eq}} = (E + E + E - E) = 2E\) and effective internal resistance of the combination is \({r_{eff}} = (r + r + r + r) = 4r\)
KCET - 2015
PHXII03:CURRENT ELECTRICITY
356866
In the electric circuit shown, each cell has an emf of \(2\,\;V\) and internal resistance is \(1 \Omega\). The external resistance is \(2\,\Omega \). The value of the current \(I\) is (in ampere)
356862
For the ladder given find \({\varepsilon _{eq}}\) and \({r_{eq}}\) across \(AB\).
1 \(4V,\,2\Omega \)
2 \(2V,\,2\Omega \)
3 \(4V,\,1\Omega \)
4 \(2V,\,6\Omega \)
Explanation:
Let \(\varepsilon \) and \(r\) be the equivalent emf and equivalent resistance across \(AB\). We can consider the circuit as Across \(CD\) also equivalent emf and resistance are \(\varepsilon \) and \(r\). \(\varepsilon ' = \frac{{\frac{\varepsilon }{r} + \frac{0}{2}}}{{\frac{1}{r} + \frac{1}{2}}} = \frac{{2\varepsilon }}{{2 + r}}\) \(\frac{1}{{r'}} = \frac{1}{r} + \frac{1}{2} \Rightarrow r' = \frac{{2r}}{{2 + r}}\) we can draw the circuit as we have taken that \(\varepsilon ' + 2 = \varepsilon \,\,{\rm{and}}\,\,r' + 1 = r\) \( \Rightarrow \frac{{2\varepsilon }}{{2 + r}} + 2 = \varepsilon \Rightarrow \varepsilon r - 4 - 2r = 0\,\,\,\,\,\left( 1 \right)\) \(r' + 1 = r \Rightarrow {r^2} - r - 2 = 0 \to \left( 2 \right)\) From (1) & (2) \(r = 2\Omega ,\quad \varepsilon = 4V\)
PHXII03:CURRENT ELECTRICITY
356863
Two batteries with e.m.f 12\(V\) and 13\(V\) are connected in parallel across a load resistor of \(10\Omega \). The internal resistances of the two batteries are \(1\Omega \,\) and \(2\Omega \) respectively. The voltage across the load lies between.
1 11.5 \(V\) and 11.6 \(V\)
2 11.4 \(V\) and 11.5 \(V\)
3 11.7 \(V\) and 11.8 \(V\)
4 11.6 \(V\) and 11.7 \(V\)
Explanation:
The equivalent emf and internal resistance are \({\varepsilon _{eq}} = \frac{{\frac{{{\varepsilon _1}}}{{{r_1}}} + \frac{{{\varepsilon _2}}}{{{r_2}}}}}{{\frac{1}{{{r_1}}} + \frac{1}{{{r_2}}}}}\) \(\frac{{\frac{{12}}{1} + \frac{{13}}{2}}}{{\frac{1}{1} + \frac{1}{2}}} = \frac{{37}}{3}V\) \({r_{eq}} = \frac{{2 \times 1}}{{2 + 1}} = \frac{2}{3}\Omega \) The equivalent circuit is \(i = \frac{{37/3}}{{10 + \frac{2}{3}}} = \frac{{37}}{{32}}\) \(\therefore {V_{10\Omega }} = i \times 10 = \frac{{37}}{{32}} \times 10 = 11.56\) volt
PHXII03:CURRENT ELECTRICITY
356864
Assertion : In order to have maximum current a series combinations of cells is used where their internal resistance is much smaller than the external resistance. Reason : Current flowing through the circuit is inversely proportional to total resistance.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
\(I=(m E / E+n r)\) In series connection, total emf of cells adds, and their internal resistances sum. When cell internal resistance \((n r)\) is much smaller than external resistance \((R)\), the current formula simplifies to approximately \(n E / R\), maximizing current with minimal impact from internal resistance. So, series connection gives maximum current. So correct option is (1).
PHXII03:CURRENT ELECTRICITY
356865
Four identical cells of emf \(E\) and internal resistance \(r\) are to be connected in series. Suppose if one of the cell is connected wrongly, the equivalent emf and effective internal resistance of the combination is
1 \(4E\,{\rm{and}}\,2r\)
2 \(2E\,{\rm{and}}\,2r\)
3 \(4E\,{\rm{and}}\,4r\)
4 \(2E\,{\rm{and}}\,4r\)
Explanation:
When 3 cells are connected correctly and one incorrectly, then equivalent emf of the combination is \({E_{eq}} = (E + E + E - E) = 2E\) and effective internal resistance of the combination is \({r_{eff}} = (r + r + r + r) = 4r\)
KCET - 2015
PHXII03:CURRENT ELECTRICITY
356866
In the electric circuit shown, each cell has an emf of \(2\,\;V\) and internal resistance is \(1 \Omega\). The external resistance is \(2\,\Omega \). The value of the current \(I\) is (in ampere)
