356971
What will be the most suitable combination of three resisotrs \(A = 2\Omega ,B = 4\Omega ,C = 6\Omega \) so that \(\left( {\frac{{22}}{3}} \right)\Omega \) is equivalent resistance of combination?
1 Parallel combination of \(A\) and \(C\) connected in series with \(B\).
2 Parallel combination of \(A\) and \(B\) connected in series with \(C\).
3 Series combination of \(A\) and \(C\) connected in parallel with \(B\).
4 Series combination of \(B\) and \(C\) connected in parallel with \(A\).
Explanation:
Required \(\frac{{22}}{7} \approx 7\Omega \) If only series, the sum \((2 + 4 + 6) > 7\). If only parallel, equivalent is less than 2. So, a mixed connection is used where \(A,B\) are in parallel and that combination is in series will \(C\). \( \Rightarrow \frac{4}{3} + 6 = \frac{{22}}{3}\)
JEE - 2022
PHXII03:CURRENT ELECTRICITY
356972
The equivalent resistance between \(A\) and \(B\) is
1 \(4.5\Omega \)
2 \(12\Omega \)
3 \(5.4\Omega \)
4 \(20\Omega \)
Explanation:
The circuit can be redrawn as \( \Rightarrow {R_{AB}} = \frac{{7 \times 3}}{{7 + 3}} + \frac{{6 \times 4}}{{6 + 4}}\) \({R_{AB}} = 4.5\,\Omega \)
PHXII03:CURRENT ELECTRICITY
356973
The equivalent resistance between \(A\) and \(B\) as shown in figure is
1 \(30\,k\Omega \)
2 \(20\,k\Omega \)
3 \(5\,k\Omega \)
4 \(10\,k\Omega \)
Explanation:
All the given resistance are connected in parallel. Hence, equivalent resistance is, \(\dfrac{1}{R_{A B}}=\dfrac{1}{20 k \Omega}+\dfrac{1}{20 k \Omega}+\dfrac{1}{10 k \Omega}\) \(=\dfrac{1+1+2}{20}=\dfrac{4}{20}=\dfrac{1}{5}\) \(R_{A B}=5 k \Omega\)
JEE - 2023
PHXII03:CURRENT ELECTRICITY
356974
A wire of resistance \(12\,\Omega \,{m^{ - 1}}\) is bent to form a complete circle of radius 10 \(cm\). The resistance between its two diametrically opposite points \(A\) and \(B\) as shown in the figure is
1 \(6\Omega \,\)
2 \(6\pi \,\Omega \)
3 \(3\,\Omega \)
4 \(0.6\pi \,\Omega \)
Explanation:
Circumference of circle \( = 2\pi r = \frac{\pi }{5}\) Resistance of wire \( = 12 \times \frac{\pi }{5} = \frac{{12\pi }}{5}\) Resistance of each section\( = \frac{{12\pi }}{{10}}\,\Omega \) \(\therefore \,\) Equivalent resistance \( = \frac{{6\pi }}{{10}} = 0.6\pi {\kern 1pt} \Omega \)
356971
What will be the most suitable combination of three resisotrs \(A = 2\Omega ,B = 4\Omega ,C = 6\Omega \) so that \(\left( {\frac{{22}}{3}} \right)\Omega \) is equivalent resistance of combination?
1 Parallel combination of \(A\) and \(C\) connected in series with \(B\).
2 Parallel combination of \(A\) and \(B\) connected in series with \(C\).
3 Series combination of \(A\) and \(C\) connected in parallel with \(B\).
4 Series combination of \(B\) and \(C\) connected in parallel with \(A\).
Explanation:
Required \(\frac{{22}}{7} \approx 7\Omega \) If only series, the sum \((2 + 4 + 6) > 7\). If only parallel, equivalent is less than 2. So, a mixed connection is used where \(A,B\) are in parallel and that combination is in series will \(C\). \( \Rightarrow \frac{4}{3} + 6 = \frac{{22}}{3}\)
JEE - 2022
PHXII03:CURRENT ELECTRICITY
356972
The equivalent resistance between \(A\) and \(B\) is
1 \(4.5\Omega \)
2 \(12\Omega \)
3 \(5.4\Omega \)
4 \(20\Omega \)
Explanation:
The circuit can be redrawn as \( \Rightarrow {R_{AB}} = \frac{{7 \times 3}}{{7 + 3}} + \frac{{6 \times 4}}{{6 + 4}}\) \({R_{AB}} = 4.5\,\Omega \)
PHXII03:CURRENT ELECTRICITY
356973
The equivalent resistance between \(A\) and \(B\) as shown in figure is
1 \(30\,k\Omega \)
2 \(20\,k\Omega \)
3 \(5\,k\Omega \)
4 \(10\,k\Omega \)
Explanation:
All the given resistance are connected in parallel. Hence, equivalent resistance is, \(\dfrac{1}{R_{A B}}=\dfrac{1}{20 k \Omega}+\dfrac{1}{20 k \Omega}+\dfrac{1}{10 k \Omega}\) \(=\dfrac{1+1+2}{20}=\dfrac{4}{20}=\dfrac{1}{5}\) \(R_{A B}=5 k \Omega\)
JEE - 2023
PHXII03:CURRENT ELECTRICITY
356974
A wire of resistance \(12\,\Omega \,{m^{ - 1}}\) is bent to form a complete circle of radius 10 \(cm\). The resistance between its two diametrically opposite points \(A\) and \(B\) as shown in the figure is
1 \(6\Omega \,\)
2 \(6\pi \,\Omega \)
3 \(3\,\Omega \)
4 \(0.6\pi \,\Omega \)
Explanation:
Circumference of circle \( = 2\pi r = \frac{\pi }{5}\) Resistance of wire \( = 12 \times \frac{\pi }{5} = \frac{{12\pi }}{5}\) Resistance of each section\( = \frac{{12\pi }}{{10}}\,\Omega \) \(\therefore \,\) Equivalent resistance \( = \frac{{6\pi }}{{10}} = 0.6\pi {\kern 1pt} \Omega \)
NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXII03:CURRENT ELECTRICITY
356971
What will be the most suitable combination of three resisotrs \(A = 2\Omega ,B = 4\Omega ,C = 6\Omega \) so that \(\left( {\frac{{22}}{3}} \right)\Omega \) is equivalent resistance of combination?
1 Parallel combination of \(A\) and \(C\) connected in series with \(B\).
2 Parallel combination of \(A\) and \(B\) connected in series with \(C\).
3 Series combination of \(A\) and \(C\) connected in parallel with \(B\).
4 Series combination of \(B\) and \(C\) connected in parallel with \(A\).
Explanation:
Required \(\frac{{22}}{7} \approx 7\Omega \) If only series, the sum \((2 + 4 + 6) > 7\). If only parallel, equivalent is less than 2. So, a mixed connection is used where \(A,B\) are in parallel and that combination is in series will \(C\). \( \Rightarrow \frac{4}{3} + 6 = \frac{{22}}{3}\)
JEE - 2022
PHXII03:CURRENT ELECTRICITY
356972
The equivalent resistance between \(A\) and \(B\) is
1 \(4.5\Omega \)
2 \(12\Omega \)
3 \(5.4\Omega \)
4 \(20\Omega \)
Explanation:
The circuit can be redrawn as \( \Rightarrow {R_{AB}} = \frac{{7 \times 3}}{{7 + 3}} + \frac{{6 \times 4}}{{6 + 4}}\) \({R_{AB}} = 4.5\,\Omega \)
PHXII03:CURRENT ELECTRICITY
356973
The equivalent resistance between \(A\) and \(B\) as shown in figure is
1 \(30\,k\Omega \)
2 \(20\,k\Omega \)
3 \(5\,k\Omega \)
4 \(10\,k\Omega \)
Explanation:
All the given resistance are connected in parallel. Hence, equivalent resistance is, \(\dfrac{1}{R_{A B}}=\dfrac{1}{20 k \Omega}+\dfrac{1}{20 k \Omega}+\dfrac{1}{10 k \Omega}\) \(=\dfrac{1+1+2}{20}=\dfrac{4}{20}=\dfrac{1}{5}\) \(R_{A B}=5 k \Omega\)
JEE - 2023
PHXII03:CURRENT ELECTRICITY
356974
A wire of resistance \(12\,\Omega \,{m^{ - 1}}\) is bent to form a complete circle of radius 10 \(cm\). The resistance between its two diametrically opposite points \(A\) and \(B\) as shown in the figure is
1 \(6\Omega \,\)
2 \(6\pi \,\Omega \)
3 \(3\,\Omega \)
4 \(0.6\pi \,\Omega \)
Explanation:
Circumference of circle \( = 2\pi r = \frac{\pi }{5}\) Resistance of wire \( = 12 \times \frac{\pi }{5} = \frac{{12\pi }}{5}\) Resistance of each section\( = \frac{{12\pi }}{{10}}\,\Omega \) \(\therefore \,\) Equivalent resistance \( = \frac{{6\pi }}{{10}} = 0.6\pi {\kern 1pt} \Omega \)
356971
What will be the most suitable combination of three resisotrs \(A = 2\Omega ,B = 4\Omega ,C = 6\Omega \) so that \(\left( {\frac{{22}}{3}} \right)\Omega \) is equivalent resistance of combination?
1 Parallel combination of \(A\) and \(C\) connected in series with \(B\).
2 Parallel combination of \(A\) and \(B\) connected in series with \(C\).
3 Series combination of \(A\) and \(C\) connected in parallel with \(B\).
4 Series combination of \(B\) and \(C\) connected in parallel with \(A\).
Explanation:
Required \(\frac{{22}}{7} \approx 7\Omega \) If only series, the sum \((2 + 4 + 6) > 7\). If only parallel, equivalent is less than 2. So, a mixed connection is used where \(A,B\) are in parallel and that combination is in series will \(C\). \( \Rightarrow \frac{4}{3} + 6 = \frac{{22}}{3}\)
JEE - 2022
PHXII03:CURRENT ELECTRICITY
356972
The equivalent resistance between \(A\) and \(B\) is
1 \(4.5\Omega \)
2 \(12\Omega \)
3 \(5.4\Omega \)
4 \(20\Omega \)
Explanation:
The circuit can be redrawn as \( \Rightarrow {R_{AB}} = \frac{{7 \times 3}}{{7 + 3}} + \frac{{6 \times 4}}{{6 + 4}}\) \({R_{AB}} = 4.5\,\Omega \)
PHXII03:CURRENT ELECTRICITY
356973
The equivalent resistance between \(A\) and \(B\) as shown in figure is
1 \(30\,k\Omega \)
2 \(20\,k\Omega \)
3 \(5\,k\Omega \)
4 \(10\,k\Omega \)
Explanation:
All the given resistance are connected in parallel. Hence, equivalent resistance is, \(\dfrac{1}{R_{A B}}=\dfrac{1}{20 k \Omega}+\dfrac{1}{20 k \Omega}+\dfrac{1}{10 k \Omega}\) \(=\dfrac{1+1+2}{20}=\dfrac{4}{20}=\dfrac{1}{5}\) \(R_{A B}=5 k \Omega\)
JEE - 2023
PHXII03:CURRENT ELECTRICITY
356974
A wire of resistance \(12\,\Omega \,{m^{ - 1}}\) is bent to form a complete circle of radius 10 \(cm\). The resistance between its two diametrically opposite points \(A\) and \(B\) as shown in the figure is
1 \(6\Omega \,\)
2 \(6\pi \,\Omega \)
3 \(3\,\Omega \)
4 \(0.6\pi \,\Omega \)
Explanation:
Circumference of circle \( = 2\pi r = \frac{\pi }{5}\) Resistance of wire \( = 12 \times \frac{\pi }{5} = \frac{{12\pi }}{5}\) Resistance of each section\( = \frac{{12\pi }}{{10}}\,\Omega \) \(\therefore \,\) Equivalent resistance \( = \frac{{6\pi }}{{10}} = 0.6\pi {\kern 1pt} \Omega \)
