Combination of Resistors
PHXII03:CURRENT ELECTRICITY

356924 Consider a spherical shell having inner radius \(a\) and outer radius \(b\). If the resistivity of the material is \(\rho \) then find the equivalent resistance between inner and outer surface.

1 \(\frac{{\rho \left( {a + b} \right)}}{{4\pi ab}}\)
2 \(\frac{{\rho \left( {b - a} \right)}}{{4\pi ab}}\)
3 \(\frac{\rho }{{4\pi b}}\)
4 \(\frac{\rho }{b}\)
PHXII03:CURRENT ELECTRICITY

356925 The equivalent resistance between the points \(A\) and \(B\) in the following circuit is
supporting img

1 \(3.12\,\Omega \)
2 \(12.48\,\Omega \)
3 \(1.56\,\Omega \)
4 \(6.24\Omega \)
PHXII03:CURRENT ELECTRICITY

356926 Equivalent resistance between the points \(A\) and \(B\) is(in \(\Omega \))
supporting img

1 \(\frac{1}{4}\)
2 \(\frac{1}{2}\)
3 \(\frac{7}{3}\)
4 \(\frac{1}{5}\)
PHXII03:CURRENT ELECTRICITY

356927 A wire of resistance \(R\) and length \(L\) is cut into 5 equal parts. If these parts are joined parallely, then resultant resistance will be :

1 \(\dfrac{1}{25} R\)
2 \(\dfrac{1}{5} R\)
3 \(5\,\,R\)
4 \(25 R\)
PHXII03:CURRENT ELECTRICITY

356924 Consider a spherical shell having inner radius \(a\) and outer radius \(b\). If the resistivity of the material is \(\rho \) then find the equivalent resistance between inner and outer surface.

1 \(\frac{{\rho \left( {a + b} \right)}}{{4\pi ab}}\)
2 \(\frac{{\rho \left( {b - a} \right)}}{{4\pi ab}}\)
3 \(\frac{\rho }{{4\pi b}}\)
4 \(\frac{\rho }{b}\)
PHXII03:CURRENT ELECTRICITY

356925 The equivalent resistance between the points \(A\) and \(B\) in the following circuit is
supporting img

1 \(3.12\,\Omega \)
2 \(12.48\,\Omega \)
3 \(1.56\,\Omega \)
4 \(6.24\Omega \)
PHXII03:CURRENT ELECTRICITY

356926 Equivalent resistance between the points \(A\) and \(B\) is(in \(\Omega \))
supporting img

1 \(\frac{1}{4}\)
2 \(\frac{1}{2}\)
3 \(\frac{7}{3}\)
4 \(\frac{1}{5}\)
PHXII03:CURRENT ELECTRICITY

356927 A wire of resistance \(R\) and length \(L\) is cut into 5 equal parts. If these parts are joined parallely, then resultant resistance will be :

1 \(\dfrac{1}{25} R\)
2 \(\dfrac{1}{5} R\)
3 \(5\,\,R\)
4 \(25 R\)
PHXII03:CURRENT ELECTRICITY

356924 Consider a spherical shell having inner radius \(a\) and outer radius \(b\). If the resistivity of the material is \(\rho \) then find the equivalent resistance between inner and outer surface.

1 \(\frac{{\rho \left( {a + b} \right)}}{{4\pi ab}}\)
2 \(\frac{{\rho \left( {b - a} \right)}}{{4\pi ab}}\)
3 \(\frac{\rho }{{4\pi b}}\)
4 \(\frac{\rho }{b}\)
PHXII03:CURRENT ELECTRICITY

356925 The equivalent resistance between the points \(A\) and \(B\) in the following circuit is
supporting img

1 \(3.12\,\Omega \)
2 \(12.48\,\Omega \)
3 \(1.56\,\Omega \)
4 \(6.24\Omega \)
PHXII03:CURRENT ELECTRICITY

356926 Equivalent resistance between the points \(A\) and \(B\) is(in \(\Omega \))
supporting img

1 \(\frac{1}{4}\)
2 \(\frac{1}{2}\)
3 \(\frac{7}{3}\)
4 \(\frac{1}{5}\)
PHXII03:CURRENT ELECTRICITY

356927 A wire of resistance \(R\) and length \(L\) is cut into 5 equal parts. If these parts are joined parallely, then resultant resistance will be :

1 \(\dfrac{1}{25} R\)
2 \(\dfrac{1}{5} R\)
3 \(5\,\,R\)
4 \(25 R\)
PHXII03:CURRENT ELECTRICITY

356924 Consider a spherical shell having inner radius \(a\) and outer radius \(b\). If the resistivity of the material is \(\rho \) then find the equivalent resistance between inner and outer surface.

1 \(\frac{{\rho \left( {a + b} \right)}}{{4\pi ab}}\)
2 \(\frac{{\rho \left( {b - a} \right)}}{{4\pi ab}}\)
3 \(\frac{\rho }{{4\pi b}}\)
4 \(\frac{\rho }{b}\)
PHXII03:CURRENT ELECTRICITY

356925 The equivalent resistance between the points \(A\) and \(B\) in the following circuit is
supporting img

1 \(3.12\,\Omega \)
2 \(12.48\,\Omega \)
3 \(1.56\,\Omega \)
4 \(6.24\Omega \)
PHXII03:CURRENT ELECTRICITY

356926 Equivalent resistance between the points \(A\) and \(B\) is(in \(\Omega \))
supporting img

1 \(\frac{1}{4}\)
2 \(\frac{1}{2}\)
3 \(\frac{7}{3}\)
4 \(\frac{1}{5}\)
PHXII03:CURRENT ELECTRICITY

356927 A wire of resistance \(R\) and length \(L\) is cut into 5 equal parts. If these parts are joined parallely, then resultant resistance will be :

1 \(\dfrac{1}{25} R\)
2 \(\dfrac{1}{5} R\)
3 \(5\,\,R\)
4 \(25 R\)