NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXII03:CURRENT ELECTRICITY
356988
The equivalent resistance between \(A\) and \(B\) is ____ .
1 \(\dfrac{2}{3} \Omega\)
2 \(\dfrac{1}{2} \Omega\)
3 \(\dfrac{3}{2} \Omega\)
4 \(\dfrac{1}{3} \Omega\)
Explanation:
The equivalent resistance is \(\dfrac{1}{R_{e q}}=\dfrac{1}{2}+\dfrac{1}{12}+\dfrac{1}{4}+\dfrac{1}{6}+\dfrac{1}{2}\) \(=\dfrac{6+1+3+2+6}{12}=\dfrac{18}{12} \dfrac{1}{R_{e q}}=\dfrac{3}{2} \Omega\) or \(R_{e q}=\dfrac{2}{3} \Omega\)
JEE - 2023
PHXII03:CURRENT ELECTRICITY
356989
The effective resistance across the points \(A\) and \(I\) is
1 \(5\,\Omega \)
2 \(1\,\Omega \)
3 \(2\,\Omega \)
4 \(0.5\,\Omega \)
Explanation:
The given circuit has perpendicular symmetry. The vertical resistors can be removed.
\({R_{AI}} = 1\Omega \)
PHXII03:CURRENT ELECTRICITY
356990
The figure shows a network of resistors each having value \(12\Omega .\) Find the equivalent resistance between points \(A\) and \(B\).
1 \(8\Omega \)
2 \(\frac{{12}}{5}\Omega \)
3 \(9\Omega \)
4 \(\frac{{11}}{3}\Omega \)
Explanation:
By applying perpendicular axis symmetry. Points lying on the line ‘\(AD\)’ have same potential therefore resistors lying on the symmetry line can be removed. The circuit can be redrawn as \({R_{AB}} = 9\Omega \)
PHXII03:CURRENT ELECTRICITY
356991
Twelve wires of equal length and same cross-section are connected in the form of a cube. If the resistance of each of the wires is \(R\) then the effective resistance between the two diagonal ends would be
1 \(2R\)
2 \(12R\)
3 \(8R\)
4 \(\frac{5}{6}R\)
Explanation:
From the symmetry it can be observed that the points \(B\), \(P\) & \(Q\) are identical w.r.to point \(A\). So the points \(B\), \(P\) & \(Q\) can be joined. Similarly the points \(C\),\(T\) & \(S\) are identical w.r.t. \(D\) and hence they can be joined. The given circuit can be simplified as follows: \(\therefore \quad {R_{AD}} = \frac{{5R}}{6}\)
356988
The equivalent resistance between \(A\) and \(B\) is ____ .
1 \(\dfrac{2}{3} \Omega\)
2 \(\dfrac{1}{2} \Omega\)
3 \(\dfrac{3}{2} \Omega\)
4 \(\dfrac{1}{3} \Omega\)
Explanation:
The equivalent resistance is \(\dfrac{1}{R_{e q}}=\dfrac{1}{2}+\dfrac{1}{12}+\dfrac{1}{4}+\dfrac{1}{6}+\dfrac{1}{2}\) \(=\dfrac{6+1+3+2+6}{12}=\dfrac{18}{12} \dfrac{1}{R_{e q}}=\dfrac{3}{2} \Omega\) or \(R_{e q}=\dfrac{2}{3} \Omega\)
JEE - 2023
PHXII03:CURRENT ELECTRICITY
356989
The effective resistance across the points \(A\) and \(I\) is
1 \(5\,\Omega \)
2 \(1\,\Omega \)
3 \(2\,\Omega \)
4 \(0.5\,\Omega \)
Explanation:
The given circuit has perpendicular symmetry. The vertical resistors can be removed.
\({R_{AI}} = 1\Omega \)
PHXII03:CURRENT ELECTRICITY
356990
The figure shows a network of resistors each having value \(12\Omega .\) Find the equivalent resistance between points \(A\) and \(B\).
1 \(8\Omega \)
2 \(\frac{{12}}{5}\Omega \)
3 \(9\Omega \)
4 \(\frac{{11}}{3}\Omega \)
Explanation:
By applying perpendicular axis symmetry. Points lying on the line ‘\(AD\)’ have same potential therefore resistors lying on the symmetry line can be removed. The circuit can be redrawn as \({R_{AB}} = 9\Omega \)
PHXII03:CURRENT ELECTRICITY
356991
Twelve wires of equal length and same cross-section are connected in the form of a cube. If the resistance of each of the wires is \(R\) then the effective resistance between the two diagonal ends would be
1 \(2R\)
2 \(12R\)
3 \(8R\)
4 \(\frac{5}{6}R\)
Explanation:
From the symmetry it can be observed that the points \(B\), \(P\) & \(Q\) are identical w.r.to point \(A\). So the points \(B\), \(P\) & \(Q\) can be joined. Similarly the points \(C\),\(T\) & \(S\) are identical w.r.t. \(D\) and hence they can be joined. The given circuit can be simplified as follows: \(\therefore \quad {R_{AD}} = \frac{{5R}}{6}\)
356988
The equivalent resistance between \(A\) and \(B\) is ____ .
1 \(\dfrac{2}{3} \Omega\)
2 \(\dfrac{1}{2} \Omega\)
3 \(\dfrac{3}{2} \Omega\)
4 \(\dfrac{1}{3} \Omega\)
Explanation:
The equivalent resistance is \(\dfrac{1}{R_{e q}}=\dfrac{1}{2}+\dfrac{1}{12}+\dfrac{1}{4}+\dfrac{1}{6}+\dfrac{1}{2}\) \(=\dfrac{6+1+3+2+6}{12}=\dfrac{18}{12} \dfrac{1}{R_{e q}}=\dfrac{3}{2} \Omega\) or \(R_{e q}=\dfrac{2}{3} \Omega\)
JEE - 2023
PHXII03:CURRENT ELECTRICITY
356989
The effective resistance across the points \(A\) and \(I\) is
1 \(5\,\Omega \)
2 \(1\,\Omega \)
3 \(2\,\Omega \)
4 \(0.5\,\Omega \)
Explanation:
The given circuit has perpendicular symmetry. The vertical resistors can be removed.
\({R_{AI}} = 1\Omega \)
PHXII03:CURRENT ELECTRICITY
356990
The figure shows a network of resistors each having value \(12\Omega .\) Find the equivalent resistance between points \(A\) and \(B\).
1 \(8\Omega \)
2 \(\frac{{12}}{5}\Omega \)
3 \(9\Omega \)
4 \(\frac{{11}}{3}\Omega \)
Explanation:
By applying perpendicular axis symmetry. Points lying on the line ‘\(AD\)’ have same potential therefore resistors lying on the symmetry line can be removed. The circuit can be redrawn as \({R_{AB}} = 9\Omega \)
PHXII03:CURRENT ELECTRICITY
356991
Twelve wires of equal length and same cross-section are connected in the form of a cube. If the resistance of each of the wires is \(R\) then the effective resistance between the two diagonal ends would be
1 \(2R\)
2 \(12R\)
3 \(8R\)
4 \(\frac{5}{6}R\)
Explanation:
From the symmetry it can be observed that the points \(B\), \(P\) & \(Q\) are identical w.r.to point \(A\). So the points \(B\), \(P\) & \(Q\) can be joined. Similarly the points \(C\),\(T\) & \(S\) are identical w.r.t. \(D\) and hence they can be joined. The given circuit can be simplified as follows: \(\therefore \quad {R_{AD}} = \frac{{5R}}{6}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
PHXII03:CURRENT ELECTRICITY
356988
The equivalent resistance between \(A\) and \(B\) is ____ .
1 \(\dfrac{2}{3} \Omega\)
2 \(\dfrac{1}{2} \Omega\)
3 \(\dfrac{3}{2} \Omega\)
4 \(\dfrac{1}{3} \Omega\)
Explanation:
The equivalent resistance is \(\dfrac{1}{R_{e q}}=\dfrac{1}{2}+\dfrac{1}{12}+\dfrac{1}{4}+\dfrac{1}{6}+\dfrac{1}{2}\) \(=\dfrac{6+1+3+2+6}{12}=\dfrac{18}{12} \dfrac{1}{R_{e q}}=\dfrac{3}{2} \Omega\) or \(R_{e q}=\dfrac{2}{3} \Omega\)
JEE - 2023
PHXII03:CURRENT ELECTRICITY
356989
The effective resistance across the points \(A\) and \(I\) is
1 \(5\,\Omega \)
2 \(1\,\Omega \)
3 \(2\,\Omega \)
4 \(0.5\,\Omega \)
Explanation:
The given circuit has perpendicular symmetry. The vertical resistors can be removed.
\({R_{AI}} = 1\Omega \)
PHXII03:CURRENT ELECTRICITY
356990
The figure shows a network of resistors each having value \(12\Omega .\) Find the equivalent resistance between points \(A\) and \(B\).
1 \(8\Omega \)
2 \(\frac{{12}}{5}\Omega \)
3 \(9\Omega \)
4 \(\frac{{11}}{3}\Omega \)
Explanation:
By applying perpendicular axis symmetry. Points lying on the line ‘\(AD\)’ have same potential therefore resistors lying on the symmetry line can be removed. The circuit can be redrawn as \({R_{AB}} = 9\Omega \)
PHXII03:CURRENT ELECTRICITY
356991
Twelve wires of equal length and same cross-section are connected in the form of a cube. If the resistance of each of the wires is \(R\) then the effective resistance between the two diagonal ends would be
1 \(2R\)
2 \(12R\)
3 \(8R\)
4 \(\frac{5}{6}R\)
Explanation:
From the symmetry it can be observed that the points \(B\), \(P\) & \(Q\) are identical w.r.to point \(A\). So the points \(B\), \(P\) & \(Q\) can be joined. Similarly the points \(C\),\(T\) & \(S\) are identical w.r.t. \(D\) and hence they can be joined. The given circuit can be simplified as follows: \(\therefore \quad {R_{AD}} = \frac{{5R}}{6}\)
