356992
The equivalent resistance between the points \(P\) and \(Q\) in the network given here is equal to (given \(r = \frac{3}{2}\Omega \))
1 \(\frac{1}{2}\Omega \)
2 \(1\Omega \)
3 \(\frac{3}{2}\Omega \)
4 \(2\Omega \)
Explanation:
The given circuit has perpendicular symmetry \({R_{PQ}} = \frac{{2r}}{3} = \frac{2}{3} \times \frac{3}{2} = 1\Omega {\rm{ }}\)
PHXII03:CURRENT ELECTRICITY
356993
Each resistance in the given cubical network has resistance of \(1\Omega \) and equivalent resistance between \(A\) and \(B\) is
1 \(\frac{5}{{12}}\Omega \)
2 \(\frac{{12}}{5}\Omega \)
3 \(\frac{5}{6}\Omega \)
4 \(\frac{6}{5}\Omega \)
Explanation:
Using symmetry of the given resistance network, it can be redrawn as 3 groups of resistances in series. The first and third group has 3 resistances in parallel and the second group has 6 resistances in parallel. \(\therefore \) Equivalent resistance between \(A\) and \(B\) \({R_{AB}} = \frac{R}{3} + \frac{R}{6} + \frac{R}{3}\) \( = \frac{5}{6}R = \frac{5}{6}\Omega \,(\because R = 1\Omega )\)
KCET - 2020
PHXII03:CURRENT ELECTRICITY
356994
The effective resistance between the points \(P\) and \(Q\) of the electrical circuit shown in the figure is
The circuit has perpendicular Symmetry. Here, by symmetry, points \(A\), \(B\), and \(C\) are at same potential, for any potential difference between \(P\) and \(Q\). The two resistors lying on \(ABC\) line can be removed. The circuit can be redrawn as \({R_{eq}} = \frac{{2Rr}}{{R + r}}\)
PHXII03:CURRENT ELECTRICITY
356995
There is an infinite wire grid with cells in the form of equilateral triangles. The resistance of each wire between neighbouring joints is \(R\). Find the net resistance of the whole grid between points \(A\) & \(B\).
1 \(R/2\)
2 \(R/3\)
3 \(R\)
4 \(R/4\)
Explanation:
We assume that current \(i\) enters at \(A\) (In the absence of \(B\)) and we distribute the current. Similarly current \(i\) comes out from \(B\) (in the absence of \(A\) ) and we distribute current. When current enters at \(A\) When current leaves from \(B\) The net current from \(A\) to \(B\) is the superposition of both the currents, i.e, \(i/6 + i/6 = i/3\) \({V_{AB}} = \left( {\frac{i}{3}} \right)\left( R \right)\) The equivalent resistance across \(A\) and \(B\) is \({{\text{R}}_{{\text{AB}}}}\) \(i{R_{AB}} = \frac{{iR}}{3}\) \({R_{AB}} = \frac{R}{3}\)
PHXII03:CURRENT ELECTRICITY
356996
Find the value of resistor to be connected between \(C\) & \(D\), so that the resistance of the entire circuit between \(A\) & \(B\) does not change with the number of elementary sets.
1 \(R\left( {\sqrt 2 - 1} \right)\)
2 \(R\left( {\sqrt 3 - 1} \right)\)
3 \(2R\)
4 \(3R\)
Explanation:
Let resistor \('x'\) be connected between \(C\) & \(D\). Even if we remove all the units, then still \({{\text{R}}_{{\text{AB}}}}\) should remain same. \( \Rightarrow {R_{AB}} = {R_{CD}} = x\) \({R_{AB}} = \frac{{R(R + x + R)}}{{R + (R + x + R)}} = \frac{{2{R^2} + Rx}}{{3R + x}}\) for \({R_{AB}}\) to be independent of no. of unit cell. \(x = \frac{{2{R^2} + Rx}}{{3R + x}}\) \({x^2} + 2Rx - 2{R^2} = 0\) \(x = R \Rightarrow x = \left( {\sqrt 3 - 1} \right)R\)
356992
The equivalent resistance between the points \(P\) and \(Q\) in the network given here is equal to (given \(r = \frac{3}{2}\Omega \))
1 \(\frac{1}{2}\Omega \)
2 \(1\Omega \)
3 \(\frac{3}{2}\Omega \)
4 \(2\Omega \)
Explanation:
The given circuit has perpendicular symmetry \({R_{PQ}} = \frac{{2r}}{3} = \frac{2}{3} \times \frac{3}{2} = 1\Omega {\rm{ }}\)
PHXII03:CURRENT ELECTRICITY
356993
Each resistance in the given cubical network has resistance of \(1\Omega \) and equivalent resistance between \(A\) and \(B\) is
1 \(\frac{5}{{12}}\Omega \)
2 \(\frac{{12}}{5}\Omega \)
3 \(\frac{5}{6}\Omega \)
4 \(\frac{6}{5}\Omega \)
Explanation:
Using symmetry of the given resistance network, it can be redrawn as 3 groups of resistances in series. The first and third group has 3 resistances in parallel and the second group has 6 resistances in parallel. \(\therefore \) Equivalent resistance between \(A\) and \(B\) \({R_{AB}} = \frac{R}{3} + \frac{R}{6} + \frac{R}{3}\) \( = \frac{5}{6}R = \frac{5}{6}\Omega \,(\because R = 1\Omega )\)
KCET - 2020
PHXII03:CURRENT ELECTRICITY
356994
The effective resistance between the points \(P\) and \(Q\) of the electrical circuit shown in the figure is
The circuit has perpendicular Symmetry. Here, by symmetry, points \(A\), \(B\), and \(C\) are at same potential, for any potential difference between \(P\) and \(Q\). The two resistors lying on \(ABC\) line can be removed. The circuit can be redrawn as \({R_{eq}} = \frac{{2Rr}}{{R + r}}\)
PHXII03:CURRENT ELECTRICITY
356995
There is an infinite wire grid with cells in the form of equilateral triangles. The resistance of each wire between neighbouring joints is \(R\). Find the net resistance of the whole grid between points \(A\) & \(B\).
1 \(R/2\)
2 \(R/3\)
3 \(R\)
4 \(R/4\)
Explanation:
We assume that current \(i\) enters at \(A\) (In the absence of \(B\)) and we distribute the current. Similarly current \(i\) comes out from \(B\) (in the absence of \(A\) ) and we distribute current. When current enters at \(A\) When current leaves from \(B\) The net current from \(A\) to \(B\) is the superposition of both the currents, i.e, \(i/6 + i/6 = i/3\) \({V_{AB}} = \left( {\frac{i}{3}} \right)\left( R \right)\) The equivalent resistance across \(A\) and \(B\) is \({{\text{R}}_{{\text{AB}}}}\) \(i{R_{AB}} = \frac{{iR}}{3}\) \({R_{AB}} = \frac{R}{3}\)
PHXII03:CURRENT ELECTRICITY
356996
Find the value of resistor to be connected between \(C\) & \(D\), so that the resistance of the entire circuit between \(A\) & \(B\) does not change with the number of elementary sets.
1 \(R\left( {\sqrt 2 - 1} \right)\)
2 \(R\left( {\sqrt 3 - 1} \right)\)
3 \(2R\)
4 \(3R\)
Explanation:
Let resistor \('x'\) be connected between \(C\) & \(D\). Even if we remove all the units, then still \({{\text{R}}_{{\text{AB}}}}\) should remain same. \( \Rightarrow {R_{AB}} = {R_{CD}} = x\) \({R_{AB}} = \frac{{R(R + x + R)}}{{R + (R + x + R)}} = \frac{{2{R^2} + Rx}}{{3R + x}}\) for \({R_{AB}}\) to be independent of no. of unit cell. \(x = \frac{{2{R^2} + Rx}}{{3R + x}}\) \({x^2} + 2Rx - 2{R^2} = 0\) \(x = R \Rightarrow x = \left( {\sqrt 3 - 1} \right)R\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXII03:CURRENT ELECTRICITY
356992
The equivalent resistance between the points \(P\) and \(Q\) in the network given here is equal to (given \(r = \frac{3}{2}\Omega \))
1 \(\frac{1}{2}\Omega \)
2 \(1\Omega \)
3 \(\frac{3}{2}\Omega \)
4 \(2\Omega \)
Explanation:
The given circuit has perpendicular symmetry \({R_{PQ}} = \frac{{2r}}{3} = \frac{2}{3} \times \frac{3}{2} = 1\Omega {\rm{ }}\)
PHXII03:CURRENT ELECTRICITY
356993
Each resistance in the given cubical network has resistance of \(1\Omega \) and equivalent resistance between \(A\) and \(B\) is
1 \(\frac{5}{{12}}\Omega \)
2 \(\frac{{12}}{5}\Omega \)
3 \(\frac{5}{6}\Omega \)
4 \(\frac{6}{5}\Omega \)
Explanation:
Using symmetry of the given resistance network, it can be redrawn as 3 groups of resistances in series. The first and third group has 3 resistances in parallel and the second group has 6 resistances in parallel. \(\therefore \) Equivalent resistance between \(A\) and \(B\) \({R_{AB}} = \frac{R}{3} + \frac{R}{6} + \frac{R}{3}\) \( = \frac{5}{6}R = \frac{5}{6}\Omega \,(\because R = 1\Omega )\)
KCET - 2020
PHXII03:CURRENT ELECTRICITY
356994
The effective resistance between the points \(P\) and \(Q\) of the electrical circuit shown in the figure is
The circuit has perpendicular Symmetry. Here, by symmetry, points \(A\), \(B\), and \(C\) are at same potential, for any potential difference between \(P\) and \(Q\). The two resistors lying on \(ABC\) line can be removed. The circuit can be redrawn as \({R_{eq}} = \frac{{2Rr}}{{R + r}}\)
PHXII03:CURRENT ELECTRICITY
356995
There is an infinite wire grid with cells in the form of equilateral triangles. The resistance of each wire between neighbouring joints is \(R\). Find the net resistance of the whole grid between points \(A\) & \(B\).
1 \(R/2\)
2 \(R/3\)
3 \(R\)
4 \(R/4\)
Explanation:
We assume that current \(i\) enters at \(A\) (In the absence of \(B\)) and we distribute the current. Similarly current \(i\) comes out from \(B\) (in the absence of \(A\) ) and we distribute current. When current enters at \(A\) When current leaves from \(B\) The net current from \(A\) to \(B\) is the superposition of both the currents, i.e, \(i/6 + i/6 = i/3\) \({V_{AB}} = \left( {\frac{i}{3}} \right)\left( R \right)\) The equivalent resistance across \(A\) and \(B\) is \({{\text{R}}_{{\text{AB}}}}\) \(i{R_{AB}} = \frac{{iR}}{3}\) \({R_{AB}} = \frac{R}{3}\)
PHXII03:CURRENT ELECTRICITY
356996
Find the value of resistor to be connected between \(C\) & \(D\), so that the resistance of the entire circuit between \(A\) & \(B\) does not change with the number of elementary sets.
1 \(R\left( {\sqrt 2 - 1} \right)\)
2 \(R\left( {\sqrt 3 - 1} \right)\)
3 \(2R\)
4 \(3R\)
Explanation:
Let resistor \('x'\) be connected between \(C\) & \(D\). Even if we remove all the units, then still \({{\text{R}}_{{\text{AB}}}}\) should remain same. \( \Rightarrow {R_{AB}} = {R_{CD}} = x\) \({R_{AB}} = \frac{{R(R + x + R)}}{{R + (R + x + R)}} = \frac{{2{R^2} + Rx}}{{3R + x}}\) for \({R_{AB}}\) to be independent of no. of unit cell. \(x = \frac{{2{R^2} + Rx}}{{3R + x}}\) \({x^2} + 2Rx - 2{R^2} = 0\) \(x = R \Rightarrow x = \left( {\sqrt 3 - 1} \right)R\)
356992
The equivalent resistance between the points \(P\) and \(Q\) in the network given here is equal to (given \(r = \frac{3}{2}\Omega \))
1 \(\frac{1}{2}\Omega \)
2 \(1\Omega \)
3 \(\frac{3}{2}\Omega \)
4 \(2\Omega \)
Explanation:
The given circuit has perpendicular symmetry \({R_{PQ}} = \frac{{2r}}{3} = \frac{2}{3} \times \frac{3}{2} = 1\Omega {\rm{ }}\)
PHXII03:CURRENT ELECTRICITY
356993
Each resistance in the given cubical network has resistance of \(1\Omega \) and equivalent resistance between \(A\) and \(B\) is
1 \(\frac{5}{{12}}\Omega \)
2 \(\frac{{12}}{5}\Omega \)
3 \(\frac{5}{6}\Omega \)
4 \(\frac{6}{5}\Omega \)
Explanation:
Using symmetry of the given resistance network, it can be redrawn as 3 groups of resistances in series. The first and third group has 3 resistances in parallel and the second group has 6 resistances in parallel. \(\therefore \) Equivalent resistance between \(A\) and \(B\) \({R_{AB}} = \frac{R}{3} + \frac{R}{6} + \frac{R}{3}\) \( = \frac{5}{6}R = \frac{5}{6}\Omega \,(\because R = 1\Omega )\)
KCET - 2020
PHXII03:CURRENT ELECTRICITY
356994
The effective resistance between the points \(P\) and \(Q\) of the electrical circuit shown in the figure is
The circuit has perpendicular Symmetry. Here, by symmetry, points \(A\), \(B\), and \(C\) are at same potential, for any potential difference between \(P\) and \(Q\). The two resistors lying on \(ABC\) line can be removed. The circuit can be redrawn as \({R_{eq}} = \frac{{2Rr}}{{R + r}}\)
PHXII03:CURRENT ELECTRICITY
356995
There is an infinite wire grid with cells in the form of equilateral triangles. The resistance of each wire between neighbouring joints is \(R\). Find the net resistance of the whole grid between points \(A\) & \(B\).
1 \(R/2\)
2 \(R/3\)
3 \(R\)
4 \(R/4\)
Explanation:
We assume that current \(i\) enters at \(A\) (In the absence of \(B\)) and we distribute the current. Similarly current \(i\) comes out from \(B\) (in the absence of \(A\) ) and we distribute current. When current enters at \(A\) When current leaves from \(B\) The net current from \(A\) to \(B\) is the superposition of both the currents, i.e, \(i/6 + i/6 = i/3\) \({V_{AB}} = \left( {\frac{i}{3}} \right)\left( R \right)\) The equivalent resistance across \(A\) and \(B\) is \({{\text{R}}_{{\text{AB}}}}\) \(i{R_{AB}} = \frac{{iR}}{3}\) \({R_{AB}} = \frac{R}{3}\)
PHXII03:CURRENT ELECTRICITY
356996
Find the value of resistor to be connected between \(C\) & \(D\), so that the resistance of the entire circuit between \(A\) & \(B\) does not change with the number of elementary sets.
1 \(R\left( {\sqrt 2 - 1} \right)\)
2 \(R\left( {\sqrt 3 - 1} \right)\)
3 \(2R\)
4 \(3R\)
Explanation:
Let resistor \('x'\) be connected between \(C\) & \(D\). Even if we remove all the units, then still \({{\text{R}}_{{\text{AB}}}}\) should remain same. \( \Rightarrow {R_{AB}} = {R_{CD}} = x\) \({R_{AB}} = \frac{{R(R + x + R)}}{{R + (R + x + R)}} = \frac{{2{R^2} + Rx}}{{3R + x}}\) for \({R_{AB}}\) to be independent of no. of unit cell. \(x = \frac{{2{R^2} + Rx}}{{3R + x}}\) \({x^2} + 2Rx - 2{R^2} = 0\) \(x = R \Rightarrow x = \left( {\sqrt 3 - 1} \right)R\)
356992
The equivalent resistance between the points \(P\) and \(Q\) in the network given here is equal to (given \(r = \frac{3}{2}\Omega \))
1 \(\frac{1}{2}\Omega \)
2 \(1\Omega \)
3 \(\frac{3}{2}\Omega \)
4 \(2\Omega \)
Explanation:
The given circuit has perpendicular symmetry \({R_{PQ}} = \frac{{2r}}{3} = \frac{2}{3} \times \frac{3}{2} = 1\Omega {\rm{ }}\)
PHXII03:CURRENT ELECTRICITY
356993
Each resistance in the given cubical network has resistance of \(1\Omega \) and equivalent resistance between \(A\) and \(B\) is
1 \(\frac{5}{{12}}\Omega \)
2 \(\frac{{12}}{5}\Omega \)
3 \(\frac{5}{6}\Omega \)
4 \(\frac{6}{5}\Omega \)
Explanation:
Using symmetry of the given resistance network, it can be redrawn as 3 groups of resistances in series. The first and third group has 3 resistances in parallel and the second group has 6 resistances in parallel. \(\therefore \) Equivalent resistance between \(A\) and \(B\) \({R_{AB}} = \frac{R}{3} + \frac{R}{6} + \frac{R}{3}\) \( = \frac{5}{6}R = \frac{5}{6}\Omega \,(\because R = 1\Omega )\)
KCET - 2020
PHXII03:CURRENT ELECTRICITY
356994
The effective resistance between the points \(P\) and \(Q\) of the electrical circuit shown in the figure is
The circuit has perpendicular Symmetry. Here, by symmetry, points \(A\), \(B\), and \(C\) are at same potential, for any potential difference between \(P\) and \(Q\). The two resistors lying on \(ABC\) line can be removed. The circuit can be redrawn as \({R_{eq}} = \frac{{2Rr}}{{R + r}}\)
PHXII03:CURRENT ELECTRICITY
356995
There is an infinite wire grid with cells in the form of equilateral triangles. The resistance of each wire between neighbouring joints is \(R\). Find the net resistance of the whole grid between points \(A\) & \(B\).
1 \(R/2\)
2 \(R/3\)
3 \(R\)
4 \(R/4\)
Explanation:
We assume that current \(i\) enters at \(A\) (In the absence of \(B\)) and we distribute the current. Similarly current \(i\) comes out from \(B\) (in the absence of \(A\) ) and we distribute current. When current enters at \(A\) When current leaves from \(B\) The net current from \(A\) to \(B\) is the superposition of both the currents, i.e, \(i/6 + i/6 = i/3\) \({V_{AB}} = \left( {\frac{i}{3}} \right)\left( R \right)\) The equivalent resistance across \(A\) and \(B\) is \({{\text{R}}_{{\text{AB}}}}\) \(i{R_{AB}} = \frac{{iR}}{3}\) \({R_{AB}} = \frac{R}{3}\)
PHXII03:CURRENT ELECTRICITY
356996
Find the value of resistor to be connected between \(C\) & \(D\), so that the resistance of the entire circuit between \(A\) & \(B\) does not change with the number of elementary sets.
1 \(R\left( {\sqrt 2 - 1} \right)\)
2 \(R\left( {\sqrt 3 - 1} \right)\)
3 \(2R\)
4 \(3R\)
Explanation:
Let resistor \('x'\) be connected between \(C\) & \(D\). Even if we remove all the units, then still \({{\text{R}}_{{\text{AB}}}}\) should remain same. \( \Rightarrow {R_{AB}} = {R_{CD}} = x\) \({R_{AB}} = \frac{{R(R + x + R)}}{{R + (R + x + R)}} = \frac{{2{R^2} + Rx}}{{3R + x}}\) for \({R_{AB}}\) to be independent of no. of unit cell. \(x = \frac{{2{R^2} + Rx}}{{3R + x}}\) \({x^2} + 2Rx - 2{R^2} = 0\) \(x = R \Rightarrow x = \left( {\sqrt 3 - 1} \right)R\)
