NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXII03:CURRENT ELECTRICITY
356928
You are given two resistances \({R_1}\,{\rm{and}}\,{R_2}\) . By using them singly, in series and in parallel, you can obtain four resistances of \(1.5\Omega ,2\Omega ,6\Omega \,{\rm{and}}\,8\Omega \). The values of \({R_1}\,{\rm{and}}\,{R_2}\) are
1 \(1\Omega ,7\Omega \)
2 \(1.5\Omega ,6.5\Omega \)
3 \(3\Omega ,5\Omega \)
4 \(2\Omega ,6\Omega \)
Explanation:
\({R_1} = 2\Omega \,{\rm{and}}\,{R_2} = 6\Omega \) In series, \(R = {R_1} + {R_2} = 8\Omega \) In parallel \(\frac{1}{R} = \frac{1}{{{R_1}}} + \frac{1}{{{R_2}}} = \frac{1}{2} + \frac{1}{6} = \frac{4}{6}\) \(\therefore \,R = \frac{6}{4} = 1.5\Omega \) \(\therefore \,\) We can get \(1.5\Omega ,2\Omega ,6\Omega \,{\rm{and}}\,8\Omega \) resistors by \(2\Omega \) and \(6\Omega \) resistors
PHXII03:CURRENT ELECTRICITY
356929
In the given figure, equivalent resistance between \(A\) and \(B\) will be
1 \(\frac{{13\,}}{{11}}\Omega \)
2 \(\frac{{14\,}}{9}\Omega \)
3 \(\frac{{1\,}}{{14}}\Omega \)
4 \(\frac{{3\,}}{{11}}\Omega \)
Explanation:
Wheatstone bridge principle is not valid for the given circuit we can apply delta star transformation.
356930
In the given circuit, the equivalent resistance between the points \(A\) and \(B\) (in ohms) is
1 9
2 11.6
3 14.5
4 21.2
Explanation:
The given circuit can be redrawn as From the above dotted region, \(3 \Omega\) and \(2 \Omega\) are in series, \(R=3+2=5 \Omega\) \(5 \Omega\) and \(6 \Omega\) are in parallel, so \(R=\dfrac{5 \times 6}{5+6}=\dfrac{30}{11} \Omega\) 30\(\dfrac{11}{11} \Omega\) and \(7 \Omega\) are in series, \(R=\dfrac{30}{11}+7=\dfrac{107}{11} \Omega\)\(9 \Omega\) \(12 \Omega, \dfrac{107}{11} \Omega\) and \(5 \Omega\) are in parallel. So, \(\dfrac{1}{R}=\dfrac{1}{12}+\dfrac{11}{107}+\dfrac{1}{5}=0.386\) \( \Rightarrow \,\,\,\,\,R = 2.6\Omega \) Hence, \(R_{\mathrm{eq}}=9+2.6=11.6 \Omega\)
PHXII03:CURRENT ELECTRICITY
356931
The equivalent resistance of the circuit shown below between points \(a\) and \(b\) is
1 \(20 \Omega\)
2 \(16 \Omega\)
3 \(3.2 \Omega\)
4 \(24 \Omega\)
Explanation:
The wheatstone bridge is balanced.So, the circuit again re-drawn as shown. Here, \(4 \Omega\) and \(4 \Omega\) are in series \(\dfrac{1}{R_{A B}}=\dfrac{1}{R_{S}}+\dfrac{1}{R_{S}^{\prime}}+\dfrac{1}{16}=\dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{16}\) \(\dfrac{1}{R_{A B}}=\dfrac{5}{16}\) \(R_{A B}=16 / 5 \Omega=3.2 \Omega\)
356928
You are given two resistances \({R_1}\,{\rm{and}}\,{R_2}\) . By using them singly, in series and in parallel, you can obtain four resistances of \(1.5\Omega ,2\Omega ,6\Omega \,{\rm{and}}\,8\Omega \). The values of \({R_1}\,{\rm{and}}\,{R_2}\) are
1 \(1\Omega ,7\Omega \)
2 \(1.5\Omega ,6.5\Omega \)
3 \(3\Omega ,5\Omega \)
4 \(2\Omega ,6\Omega \)
Explanation:
\({R_1} = 2\Omega \,{\rm{and}}\,{R_2} = 6\Omega \) In series, \(R = {R_1} + {R_2} = 8\Omega \) In parallel \(\frac{1}{R} = \frac{1}{{{R_1}}} + \frac{1}{{{R_2}}} = \frac{1}{2} + \frac{1}{6} = \frac{4}{6}\) \(\therefore \,R = \frac{6}{4} = 1.5\Omega \) \(\therefore \,\) We can get \(1.5\Omega ,2\Omega ,6\Omega \,{\rm{and}}\,8\Omega \) resistors by \(2\Omega \) and \(6\Omega \) resistors
PHXII03:CURRENT ELECTRICITY
356929
In the given figure, equivalent resistance between \(A\) and \(B\) will be
1 \(\frac{{13\,}}{{11}}\Omega \)
2 \(\frac{{14\,}}{9}\Omega \)
3 \(\frac{{1\,}}{{14}}\Omega \)
4 \(\frac{{3\,}}{{11}}\Omega \)
Explanation:
Wheatstone bridge principle is not valid for the given circuit we can apply delta star transformation.
356930
In the given circuit, the equivalent resistance between the points \(A\) and \(B\) (in ohms) is
1 9
2 11.6
3 14.5
4 21.2
Explanation:
The given circuit can be redrawn as From the above dotted region, \(3 \Omega\) and \(2 \Omega\) are in series, \(R=3+2=5 \Omega\) \(5 \Omega\) and \(6 \Omega\) are in parallel, so \(R=\dfrac{5 \times 6}{5+6}=\dfrac{30}{11} \Omega\) 30\(\dfrac{11}{11} \Omega\) and \(7 \Omega\) are in series, \(R=\dfrac{30}{11}+7=\dfrac{107}{11} \Omega\)\(9 \Omega\) \(12 \Omega, \dfrac{107}{11} \Omega\) and \(5 \Omega\) are in parallel. So, \(\dfrac{1}{R}=\dfrac{1}{12}+\dfrac{11}{107}+\dfrac{1}{5}=0.386\) \( \Rightarrow \,\,\,\,\,R = 2.6\Omega \) Hence, \(R_{\mathrm{eq}}=9+2.6=11.6 \Omega\)
PHXII03:CURRENT ELECTRICITY
356931
The equivalent resistance of the circuit shown below between points \(a\) and \(b\) is
1 \(20 \Omega\)
2 \(16 \Omega\)
3 \(3.2 \Omega\)
4 \(24 \Omega\)
Explanation:
The wheatstone bridge is balanced.So, the circuit again re-drawn as shown. Here, \(4 \Omega\) and \(4 \Omega\) are in series \(\dfrac{1}{R_{A B}}=\dfrac{1}{R_{S}}+\dfrac{1}{R_{S}^{\prime}}+\dfrac{1}{16}=\dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{16}\) \(\dfrac{1}{R_{A B}}=\dfrac{5}{16}\) \(R_{A B}=16 / 5 \Omega=3.2 \Omega\)
356928
You are given two resistances \({R_1}\,{\rm{and}}\,{R_2}\) . By using them singly, in series and in parallel, you can obtain four resistances of \(1.5\Omega ,2\Omega ,6\Omega \,{\rm{and}}\,8\Omega \). The values of \({R_1}\,{\rm{and}}\,{R_2}\) are
1 \(1\Omega ,7\Omega \)
2 \(1.5\Omega ,6.5\Omega \)
3 \(3\Omega ,5\Omega \)
4 \(2\Omega ,6\Omega \)
Explanation:
\({R_1} = 2\Omega \,{\rm{and}}\,{R_2} = 6\Omega \) In series, \(R = {R_1} + {R_2} = 8\Omega \) In parallel \(\frac{1}{R} = \frac{1}{{{R_1}}} + \frac{1}{{{R_2}}} = \frac{1}{2} + \frac{1}{6} = \frac{4}{6}\) \(\therefore \,R = \frac{6}{4} = 1.5\Omega \) \(\therefore \,\) We can get \(1.5\Omega ,2\Omega ,6\Omega \,{\rm{and}}\,8\Omega \) resistors by \(2\Omega \) and \(6\Omega \) resistors
PHXII03:CURRENT ELECTRICITY
356929
In the given figure, equivalent resistance between \(A\) and \(B\) will be
1 \(\frac{{13\,}}{{11}}\Omega \)
2 \(\frac{{14\,}}{9}\Omega \)
3 \(\frac{{1\,}}{{14}}\Omega \)
4 \(\frac{{3\,}}{{11}}\Omega \)
Explanation:
Wheatstone bridge principle is not valid for the given circuit we can apply delta star transformation.
356930
In the given circuit, the equivalent resistance between the points \(A\) and \(B\) (in ohms) is
1 9
2 11.6
3 14.5
4 21.2
Explanation:
The given circuit can be redrawn as From the above dotted region, \(3 \Omega\) and \(2 \Omega\) are in series, \(R=3+2=5 \Omega\) \(5 \Omega\) and \(6 \Omega\) are in parallel, so \(R=\dfrac{5 \times 6}{5+6}=\dfrac{30}{11} \Omega\) 30\(\dfrac{11}{11} \Omega\) and \(7 \Omega\) are in series, \(R=\dfrac{30}{11}+7=\dfrac{107}{11} \Omega\)\(9 \Omega\) \(12 \Omega, \dfrac{107}{11} \Omega\) and \(5 \Omega\) are in parallel. So, \(\dfrac{1}{R}=\dfrac{1}{12}+\dfrac{11}{107}+\dfrac{1}{5}=0.386\) \( \Rightarrow \,\,\,\,\,R = 2.6\Omega \) Hence, \(R_{\mathrm{eq}}=9+2.6=11.6 \Omega\)
PHXII03:CURRENT ELECTRICITY
356931
The equivalent resistance of the circuit shown below between points \(a\) and \(b\) is
1 \(20 \Omega\)
2 \(16 \Omega\)
3 \(3.2 \Omega\)
4 \(24 \Omega\)
Explanation:
The wheatstone bridge is balanced.So, the circuit again re-drawn as shown. Here, \(4 \Omega\) and \(4 \Omega\) are in series \(\dfrac{1}{R_{A B}}=\dfrac{1}{R_{S}}+\dfrac{1}{R_{S}^{\prime}}+\dfrac{1}{16}=\dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{16}\) \(\dfrac{1}{R_{A B}}=\dfrac{5}{16}\) \(R_{A B}=16 / 5 \Omega=3.2 \Omega\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
PHXII03:CURRENT ELECTRICITY
356928
You are given two resistances \({R_1}\,{\rm{and}}\,{R_2}\) . By using them singly, in series and in parallel, you can obtain four resistances of \(1.5\Omega ,2\Omega ,6\Omega \,{\rm{and}}\,8\Omega \). The values of \({R_1}\,{\rm{and}}\,{R_2}\) are
1 \(1\Omega ,7\Omega \)
2 \(1.5\Omega ,6.5\Omega \)
3 \(3\Omega ,5\Omega \)
4 \(2\Omega ,6\Omega \)
Explanation:
\({R_1} = 2\Omega \,{\rm{and}}\,{R_2} = 6\Omega \) In series, \(R = {R_1} + {R_2} = 8\Omega \) In parallel \(\frac{1}{R} = \frac{1}{{{R_1}}} + \frac{1}{{{R_2}}} = \frac{1}{2} + \frac{1}{6} = \frac{4}{6}\) \(\therefore \,R = \frac{6}{4} = 1.5\Omega \) \(\therefore \,\) We can get \(1.5\Omega ,2\Omega ,6\Omega \,{\rm{and}}\,8\Omega \) resistors by \(2\Omega \) and \(6\Omega \) resistors
PHXII03:CURRENT ELECTRICITY
356929
In the given figure, equivalent resistance between \(A\) and \(B\) will be
1 \(\frac{{13\,}}{{11}}\Omega \)
2 \(\frac{{14\,}}{9}\Omega \)
3 \(\frac{{1\,}}{{14}}\Omega \)
4 \(\frac{{3\,}}{{11}}\Omega \)
Explanation:
Wheatstone bridge principle is not valid for the given circuit we can apply delta star transformation.
356930
In the given circuit, the equivalent resistance between the points \(A\) and \(B\) (in ohms) is
1 9
2 11.6
3 14.5
4 21.2
Explanation:
The given circuit can be redrawn as From the above dotted region, \(3 \Omega\) and \(2 \Omega\) are in series, \(R=3+2=5 \Omega\) \(5 \Omega\) and \(6 \Omega\) are in parallel, so \(R=\dfrac{5 \times 6}{5+6}=\dfrac{30}{11} \Omega\) 30\(\dfrac{11}{11} \Omega\) and \(7 \Omega\) are in series, \(R=\dfrac{30}{11}+7=\dfrac{107}{11} \Omega\)\(9 \Omega\) \(12 \Omega, \dfrac{107}{11} \Omega\) and \(5 \Omega\) are in parallel. So, \(\dfrac{1}{R}=\dfrac{1}{12}+\dfrac{11}{107}+\dfrac{1}{5}=0.386\) \( \Rightarrow \,\,\,\,\,R = 2.6\Omega \) Hence, \(R_{\mathrm{eq}}=9+2.6=11.6 \Omega\)
PHXII03:CURRENT ELECTRICITY
356931
The equivalent resistance of the circuit shown below between points \(a\) and \(b\) is
1 \(20 \Omega\)
2 \(16 \Omega\)
3 \(3.2 \Omega\)
4 \(24 \Omega\)
Explanation:
The wheatstone bridge is balanced.So, the circuit again re-drawn as shown. Here, \(4 \Omega\) and \(4 \Omega\) are in series \(\dfrac{1}{R_{A B}}=\dfrac{1}{R_{S}}+\dfrac{1}{R_{S}^{\prime}}+\dfrac{1}{16}=\dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{16}\) \(\dfrac{1}{R_{A B}}=\dfrac{5}{16}\) \(R_{A B}=16 / 5 \Omega=3.2 \Omega\)
