356975
The effective resistance between \(p\) and \(q\) in given figure is
1 \(2\,\Omega \)
2 \(3\,\Omega \)
3 \(5\,\Omega \)
4 \(6\,\Omega \)
Explanation:
Figure is equivalent to the one shown below. It is a Wheatstone's bridge in which \(P=\dfrac{6 \times 3}{6+3}=2 \Omega\) \(Q=\dfrac{8 \times 8}{8+8}=\dfrac{64}{16}=4 \Omega\) \(R=\dfrac{4 \times 4}{4+4}=\dfrac{16}{8}=2 \Omega\) and \(S=\dfrac{20 \times 5}{20+5}=4 \Omega\) We find that, \(\dfrac{P}{Q}=\dfrac{2}{4}=\dfrac{R}{S}\) \(i.e.\) the bridge is balanced and resistance of arm \(C D\) is ineffective. Effective resistance between \(p\) and \(q\) \(\begin{aligned}& R_{A B}=\dfrac{(P+Q)(R+S)}{P+Q+R+S} \\& =\dfrac{(2+4)(2+4)}{2+4+2+4}=\dfrac{6 \times 6}{12}=3 \Omega\end{aligned}\) So, correct option is (2)
AIIMS - 2018
PHXII03:CURRENT ELECTRICITY
356976
The equivalent resistance between points \(a\) and \(b\) of a network shown in the figure is given by
1 \(\frac{5}{4}R\)
2 \(\frac{4}{3}R\)
3 \(\frac{4}{5}R\)
4 \(\frac{3}{4}R\)
Explanation:
The circuit can be redrawn as \({R_{eq}} = R + R/3 = 4R/3\)
PHXII03:CURRENT ELECTRICITY
356977
A uniform wire of resistance \(R\) stretched uniformly to \(n\) times and then cut to from five wires of equal length. These wires are arranged as shown in the figure. The effective resistance between points \(A\) and \(B\) is
1 \(R/\left( {5{n^2}} \right)\)
2 \(R/\left( {5n} \right)\)
3 \({n^2}R/5\)
4 \(nR/5\)
Explanation:
\(R = \frac{{\rho {{\rm{l}}_1}}}{{{A_1}}}\) [\(\rho = \) resistivity of the wire] \({{\rm{l}}_2} = n{{\rm{l}}_1},{A_1}{{\rm{l}}_1} = {A_2}{{\rm{l}}_2}\) \( \Rightarrow {A_2} = {A_1}/n\) New resistance of the wire \( = R' = \frac{{\rho {{\rm{l}}_2}}}{{{A_2}}} = {n^2}R\) Resistance of each wire is \({n^2}R/5\) Since the wires are arranged to form balanced Wheatstone’s bridge \({R_{AB}} = \frac{{\left( {{n^2}R/5} \right) + \left( {{n^2}R/5} \right)}}{2} = \frac{{{n^2}R}}{5}\)
PHXII03:CURRENT ELECTRICITY
356978
In the given circuit all resistances are of value of \(R\) ohm each. The equivalent resistance between \(A\) and \(B\) is:
1 \(\frac{{5R}}{2}\)
2 \(3R\)
3 \(\frac{{5R}}{3}\)
4 \(2R\)
Explanation:
If we combine or squeeze all the points \(C\) then the circuit will be reduced to
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PHXII03:CURRENT ELECTRICITY
356975
The effective resistance between \(p\) and \(q\) in given figure is
1 \(2\,\Omega \)
2 \(3\,\Omega \)
3 \(5\,\Omega \)
4 \(6\,\Omega \)
Explanation:
Figure is equivalent to the one shown below. It is a Wheatstone's bridge in which \(P=\dfrac{6 \times 3}{6+3}=2 \Omega\) \(Q=\dfrac{8 \times 8}{8+8}=\dfrac{64}{16}=4 \Omega\) \(R=\dfrac{4 \times 4}{4+4}=\dfrac{16}{8}=2 \Omega\) and \(S=\dfrac{20 \times 5}{20+5}=4 \Omega\) We find that, \(\dfrac{P}{Q}=\dfrac{2}{4}=\dfrac{R}{S}\) \(i.e.\) the bridge is balanced and resistance of arm \(C D\) is ineffective. Effective resistance between \(p\) and \(q\) \(\begin{aligned}& R_{A B}=\dfrac{(P+Q)(R+S)}{P+Q+R+S} \\& =\dfrac{(2+4)(2+4)}{2+4+2+4}=\dfrac{6 \times 6}{12}=3 \Omega\end{aligned}\) So, correct option is (2)
AIIMS - 2018
PHXII03:CURRENT ELECTRICITY
356976
The equivalent resistance between points \(a\) and \(b\) of a network shown in the figure is given by
1 \(\frac{5}{4}R\)
2 \(\frac{4}{3}R\)
3 \(\frac{4}{5}R\)
4 \(\frac{3}{4}R\)
Explanation:
The circuit can be redrawn as \({R_{eq}} = R + R/3 = 4R/3\)
PHXII03:CURRENT ELECTRICITY
356977
A uniform wire of resistance \(R\) stretched uniformly to \(n\) times and then cut to from five wires of equal length. These wires are arranged as shown in the figure. The effective resistance between points \(A\) and \(B\) is
1 \(R/\left( {5{n^2}} \right)\)
2 \(R/\left( {5n} \right)\)
3 \({n^2}R/5\)
4 \(nR/5\)
Explanation:
\(R = \frac{{\rho {{\rm{l}}_1}}}{{{A_1}}}\) [\(\rho = \) resistivity of the wire] \({{\rm{l}}_2} = n{{\rm{l}}_1},{A_1}{{\rm{l}}_1} = {A_2}{{\rm{l}}_2}\) \( \Rightarrow {A_2} = {A_1}/n\) New resistance of the wire \( = R' = \frac{{\rho {{\rm{l}}_2}}}{{{A_2}}} = {n^2}R\) Resistance of each wire is \({n^2}R/5\) Since the wires are arranged to form balanced Wheatstone’s bridge \({R_{AB}} = \frac{{\left( {{n^2}R/5} \right) + \left( {{n^2}R/5} \right)}}{2} = \frac{{{n^2}R}}{5}\)
PHXII03:CURRENT ELECTRICITY
356978
In the given circuit all resistances are of value of \(R\) ohm each. The equivalent resistance between \(A\) and \(B\) is:
1 \(\frac{{5R}}{2}\)
2 \(3R\)
3 \(\frac{{5R}}{3}\)
4 \(2R\)
Explanation:
If we combine or squeeze all the points \(C\) then the circuit will be reduced to
356975
The effective resistance between \(p\) and \(q\) in given figure is
1 \(2\,\Omega \)
2 \(3\,\Omega \)
3 \(5\,\Omega \)
4 \(6\,\Omega \)
Explanation:
Figure is equivalent to the one shown below. It is a Wheatstone's bridge in which \(P=\dfrac{6 \times 3}{6+3}=2 \Omega\) \(Q=\dfrac{8 \times 8}{8+8}=\dfrac{64}{16}=4 \Omega\) \(R=\dfrac{4 \times 4}{4+4}=\dfrac{16}{8}=2 \Omega\) and \(S=\dfrac{20 \times 5}{20+5}=4 \Omega\) We find that, \(\dfrac{P}{Q}=\dfrac{2}{4}=\dfrac{R}{S}\) \(i.e.\) the bridge is balanced and resistance of arm \(C D\) is ineffective. Effective resistance between \(p\) and \(q\) \(\begin{aligned}& R_{A B}=\dfrac{(P+Q)(R+S)}{P+Q+R+S} \\& =\dfrac{(2+4)(2+4)}{2+4+2+4}=\dfrac{6 \times 6}{12}=3 \Omega\end{aligned}\) So, correct option is (2)
AIIMS - 2018
PHXII03:CURRENT ELECTRICITY
356976
The equivalent resistance between points \(a\) and \(b\) of a network shown in the figure is given by
1 \(\frac{5}{4}R\)
2 \(\frac{4}{3}R\)
3 \(\frac{4}{5}R\)
4 \(\frac{3}{4}R\)
Explanation:
The circuit can be redrawn as \({R_{eq}} = R + R/3 = 4R/3\)
PHXII03:CURRENT ELECTRICITY
356977
A uniform wire of resistance \(R\) stretched uniformly to \(n\) times and then cut to from five wires of equal length. These wires are arranged as shown in the figure. The effective resistance between points \(A\) and \(B\) is
1 \(R/\left( {5{n^2}} \right)\)
2 \(R/\left( {5n} \right)\)
3 \({n^2}R/5\)
4 \(nR/5\)
Explanation:
\(R = \frac{{\rho {{\rm{l}}_1}}}{{{A_1}}}\) [\(\rho = \) resistivity of the wire] \({{\rm{l}}_2} = n{{\rm{l}}_1},{A_1}{{\rm{l}}_1} = {A_2}{{\rm{l}}_2}\) \( \Rightarrow {A_2} = {A_1}/n\) New resistance of the wire \( = R' = \frac{{\rho {{\rm{l}}_2}}}{{{A_2}}} = {n^2}R\) Resistance of each wire is \({n^2}R/5\) Since the wires are arranged to form balanced Wheatstone’s bridge \({R_{AB}} = \frac{{\left( {{n^2}R/5} \right) + \left( {{n^2}R/5} \right)}}{2} = \frac{{{n^2}R}}{5}\)
PHXII03:CURRENT ELECTRICITY
356978
In the given circuit all resistances are of value of \(R\) ohm each. The equivalent resistance between \(A\) and \(B\) is:
1 \(\frac{{5R}}{2}\)
2 \(3R\)
3 \(\frac{{5R}}{3}\)
4 \(2R\)
Explanation:
If we combine or squeeze all the points \(C\) then the circuit will be reduced to
356975
The effective resistance between \(p\) and \(q\) in given figure is
1 \(2\,\Omega \)
2 \(3\,\Omega \)
3 \(5\,\Omega \)
4 \(6\,\Omega \)
Explanation:
Figure is equivalent to the one shown below. It is a Wheatstone's bridge in which \(P=\dfrac{6 \times 3}{6+3}=2 \Omega\) \(Q=\dfrac{8 \times 8}{8+8}=\dfrac{64}{16}=4 \Omega\) \(R=\dfrac{4 \times 4}{4+4}=\dfrac{16}{8}=2 \Omega\) and \(S=\dfrac{20 \times 5}{20+5}=4 \Omega\) We find that, \(\dfrac{P}{Q}=\dfrac{2}{4}=\dfrac{R}{S}\) \(i.e.\) the bridge is balanced and resistance of arm \(C D\) is ineffective. Effective resistance between \(p\) and \(q\) \(\begin{aligned}& R_{A B}=\dfrac{(P+Q)(R+S)}{P+Q+R+S} \\& =\dfrac{(2+4)(2+4)}{2+4+2+4}=\dfrac{6 \times 6}{12}=3 \Omega\end{aligned}\) So, correct option is (2)
AIIMS - 2018
PHXII03:CURRENT ELECTRICITY
356976
The equivalent resistance between points \(a\) and \(b\) of a network shown in the figure is given by
1 \(\frac{5}{4}R\)
2 \(\frac{4}{3}R\)
3 \(\frac{4}{5}R\)
4 \(\frac{3}{4}R\)
Explanation:
The circuit can be redrawn as \({R_{eq}} = R + R/3 = 4R/3\)
PHXII03:CURRENT ELECTRICITY
356977
A uniform wire of resistance \(R\) stretched uniformly to \(n\) times and then cut to from five wires of equal length. These wires are arranged as shown in the figure. The effective resistance between points \(A\) and \(B\) is
1 \(R/\left( {5{n^2}} \right)\)
2 \(R/\left( {5n} \right)\)
3 \({n^2}R/5\)
4 \(nR/5\)
Explanation:
\(R = \frac{{\rho {{\rm{l}}_1}}}{{{A_1}}}\) [\(\rho = \) resistivity of the wire] \({{\rm{l}}_2} = n{{\rm{l}}_1},{A_1}{{\rm{l}}_1} = {A_2}{{\rm{l}}_2}\) \( \Rightarrow {A_2} = {A_1}/n\) New resistance of the wire \( = R' = \frac{{\rho {{\rm{l}}_2}}}{{{A_2}}} = {n^2}R\) Resistance of each wire is \({n^2}R/5\) Since the wires are arranged to form balanced Wheatstone’s bridge \({R_{AB}} = \frac{{\left( {{n^2}R/5} \right) + \left( {{n^2}R/5} \right)}}{2} = \frac{{{n^2}R}}{5}\)
PHXII03:CURRENT ELECTRICITY
356978
In the given circuit all resistances are of value of \(R\) ohm each. The equivalent resistance between \(A\) and \(B\) is:
1 \(\frac{{5R}}{2}\)
2 \(3R\)
3 \(\frac{{5R}}{3}\)
4 \(2R\)
Explanation:
If we combine or squeeze all the points \(C\) then the circuit will be reduced to
