356959
If the equivalent resistance between points \({A}\) and \({B}\) is \({R_{A B}=\dfrac{45}{N} \Omega}\). Find the value of \({N}\) is
1 4
2 7
3 9
4 2
Explanation:
The figure can be redrawn as \({R_{AB}} = 22.5\,\Omega \) \({R_{AB}} = \frac{{45}}{N}\Omega = 22.5\) \(\therefore \,\,\,N = \frac{{45}}{{22.5}} = 2\) \( = 2\) So answer is 2.
PHXII03:CURRENT ELECTRICITY
356960
A \(5\Omega \) resistance wire is bent to form a ring. Find the resistance across the diameter of the wire
1 \(0.625\Omega \)
2 \(1.25\Omega \)
3 \(2.5\Omega \)
4 \(5\Omega \)
Explanation:
The resistance of each semicircular portion is \(2.5\Omega .\)The total resistance across the diameter is \(1.25\Omega .\)
PHXII03:CURRENT ELECTRICITY
356961
The effective resistance between \(P\) and \(Q\) for the following network is
1 \(\frac{1}{{12}}\Omega \)
2 \({\rm{21\Omega }}\)
3 \({\rm{12\Omega }}\)
4 \(\frac{1}{{21}}\Omega \)
Explanation:
In the given circuit, two \(3\,\Omega \) resistors are in series and this combination is in parallel with a 6\(W\) resistor . Their equivalent resistance \( = \frac{{\left( {3 + 3} \right) \times 6}}{{\left( {3 + 3} \right) + 6}}\Omega = \frac{{36}}{{12}}\Omega = 3\,\Omega \) This resistance is in series with the \(4\Omega \) and \(5\Omega \) resistances. So the effective resistance between \(P\) and \(Q\) is \({R_{PQ}} = 4\Omega + 3\Omega + 5\Omega = 12\Omega \)
356959
If the equivalent resistance between points \({A}\) and \({B}\) is \({R_{A B}=\dfrac{45}{N} \Omega}\). Find the value of \({N}\) is
1 4
2 7
3 9
4 2
Explanation:
The figure can be redrawn as \({R_{AB}} = 22.5\,\Omega \) \({R_{AB}} = \frac{{45}}{N}\Omega = 22.5\) \(\therefore \,\,\,N = \frac{{45}}{{22.5}} = 2\) \( = 2\) So answer is 2.
PHXII03:CURRENT ELECTRICITY
356960
A \(5\Omega \) resistance wire is bent to form a ring. Find the resistance across the diameter of the wire
1 \(0.625\Omega \)
2 \(1.25\Omega \)
3 \(2.5\Omega \)
4 \(5\Omega \)
Explanation:
The resistance of each semicircular portion is \(2.5\Omega .\)The total resistance across the diameter is \(1.25\Omega .\)
PHXII03:CURRENT ELECTRICITY
356961
The effective resistance between \(P\) and \(Q\) for the following network is
1 \(\frac{1}{{12}}\Omega \)
2 \({\rm{21\Omega }}\)
3 \({\rm{12\Omega }}\)
4 \(\frac{1}{{21}}\Omega \)
Explanation:
In the given circuit, two \(3\,\Omega \) resistors are in series and this combination is in parallel with a 6\(W\) resistor . Their equivalent resistance \( = \frac{{\left( {3 + 3} \right) \times 6}}{{\left( {3 + 3} \right) + 6}}\Omega = \frac{{36}}{{12}}\Omega = 3\,\Omega \) This resistance is in series with the \(4\Omega \) and \(5\Omega \) resistances. So the effective resistance between \(P\) and \(Q\) is \({R_{PQ}} = 4\Omega + 3\Omega + 5\Omega = 12\Omega \)
356959
If the equivalent resistance between points \({A}\) and \({B}\) is \({R_{A B}=\dfrac{45}{N} \Omega}\). Find the value of \({N}\) is
1 4
2 7
3 9
4 2
Explanation:
The figure can be redrawn as \({R_{AB}} = 22.5\,\Omega \) \({R_{AB}} = \frac{{45}}{N}\Omega = 22.5\) \(\therefore \,\,\,N = \frac{{45}}{{22.5}} = 2\) \( = 2\) So answer is 2.
PHXII03:CURRENT ELECTRICITY
356960
A \(5\Omega \) resistance wire is bent to form a ring. Find the resistance across the diameter of the wire
1 \(0.625\Omega \)
2 \(1.25\Omega \)
3 \(2.5\Omega \)
4 \(5\Omega \)
Explanation:
The resistance of each semicircular portion is \(2.5\Omega .\)The total resistance across the diameter is \(1.25\Omega .\)
PHXII03:CURRENT ELECTRICITY
356961
The effective resistance between \(P\) and \(Q\) for the following network is
1 \(\frac{1}{{12}}\Omega \)
2 \({\rm{21\Omega }}\)
3 \({\rm{12\Omega }}\)
4 \(\frac{1}{{21}}\Omega \)
Explanation:
In the given circuit, two \(3\,\Omega \) resistors are in series and this combination is in parallel with a 6\(W\) resistor . Their equivalent resistance \( = \frac{{\left( {3 + 3} \right) \times 6}}{{\left( {3 + 3} \right) + 6}}\Omega = \frac{{36}}{{12}}\Omega = 3\,\Omega \) This resistance is in series with the \(4\Omega \) and \(5\Omega \) resistances. So the effective resistance between \(P\) and \(Q\) is \({R_{PQ}} = 4\Omega + 3\Omega + 5\Omega = 12\Omega \)
356959
If the equivalent resistance between points \({A}\) and \({B}\) is \({R_{A B}=\dfrac{45}{N} \Omega}\). Find the value of \({N}\) is
1 4
2 7
3 9
4 2
Explanation:
The figure can be redrawn as \({R_{AB}} = 22.5\,\Omega \) \({R_{AB}} = \frac{{45}}{N}\Omega = 22.5\) \(\therefore \,\,\,N = \frac{{45}}{{22.5}} = 2\) \( = 2\) So answer is 2.
PHXII03:CURRENT ELECTRICITY
356960
A \(5\Omega \) resistance wire is bent to form a ring. Find the resistance across the diameter of the wire
1 \(0.625\Omega \)
2 \(1.25\Omega \)
3 \(2.5\Omega \)
4 \(5\Omega \)
Explanation:
The resistance of each semicircular portion is \(2.5\Omega .\)The total resistance across the diameter is \(1.25\Omega .\)
PHXII03:CURRENT ELECTRICITY
356961
The effective resistance between \(P\) and \(Q\) for the following network is
1 \(\frac{1}{{12}}\Omega \)
2 \({\rm{21\Omega }}\)
3 \({\rm{12\Omega }}\)
4 \(\frac{1}{{21}}\Omega \)
Explanation:
In the given circuit, two \(3\,\Omega \) resistors are in series and this combination is in parallel with a 6\(W\) resistor . Their equivalent resistance \( = \frac{{\left( {3 + 3} \right) \times 6}}{{\left( {3 + 3} \right) + 6}}\Omega = \frac{{36}}{{12}}\Omega = 3\,\Omega \) This resistance is in series with the \(4\Omega \) and \(5\Omega \) resistances. So the effective resistance between \(P\) and \(Q\) is \({R_{PQ}} = 4\Omega + 3\Omega + 5\Omega = 12\Omega \)
