NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXII03:CURRENT ELECTRICITY
357418
If the galvanometer \(G\) does not show any deflection in the circuit shown, the value of \(R\) is given by:
1 \(50\,\Omega \)
2 \(100\,\Omega \)
3 \(400\,\Omega \)
4 \(200\,\Omega \)
Explanation:
For no reading galvanometer,\({V_{AB}} = {V_{CD}}\). \(\left( {\frac{{{E_1} - {E_2}}}{{400}}} \right) = \frac{{{E_2}}}{R}\) \(\left( {\frac{{10 - 2}}{{400}}} \right) = \frac{2}{R} \Rightarrow R = 100\,\Omega \).
NEET - 2023
PHXII03:CURRENT ELECTRICITY
357419
For the value of \(R\) in the circuit as shown in figure, current passing through \(4\Omega \) resistance will be zero.
1 \(4\Omega \)
2 \(2\Omega \)
3 \(3\Omega \)
4 \(1\Omega \)
Explanation:
Since, current does not flow in the \(4\Omega \) resistance, hence resistance \(4\Omega \) becomes ineffective in current. Current through resistance \(2\Omega \) is \(i = \frac{{9 - 6}}{2} = \frac{3}{2}A\) In circuit \(ABCDEFA\), \(9 - 3 - Ri - 2i = 0\) \( \Rightarrow R = 2\Omega \)
PHXII03:CURRENT ELECTRICITY
357420
Consider the following two statements: I. Kirchoff’s junction law follows from the conservation of charge. II. Kirchoff’s loop law follows from the conservation of energy. Which of the following is correct?
1 Both I and II are correct
2 I is wrong and II is correct
3 I is correct and II is wrong
4 Both I and II are wrong
Explanation:
Conceptual Question
PHXII03:CURRENT ELECTRICITY
357421
A \(100\,V\) voltmeter of internal resistance \(20\,k\,\Omega \) in series with a high resistance \(R\) is connected to \(110\,V\) line. The voltmeter reads \(5\,V\), the value of \(R\) is
1 \(210\,k\,\Omega \)
2 \(315\,k\,\Omega \)
3 \(420\,k\,\Omega \)
4 \(4440\,k\,\Omega \)
Explanation:
The circuit is as shown figure Potential difference across voltmeter \( = 5\;V\) \(\therefore\) Current in the circuit, \(i = \frac{5}{{{R_V}}} = \frac{5}{{20 \times {{10}^3}}} = 0.25 \times {10^{ - 3}}\;A\) Voltage across \(R,{V_1} = 110 - 5 = 105\;V\) Hence, \(R=\dfrac{V_{1}}{i}=\dfrac{105}{0.25 \times 10^{-3}}=420 \times 10^{3}=420 \mathrm{k} \Omega\) So, correct option is (3).
357418
If the galvanometer \(G\) does not show any deflection in the circuit shown, the value of \(R\) is given by:
1 \(50\,\Omega \)
2 \(100\,\Omega \)
3 \(400\,\Omega \)
4 \(200\,\Omega \)
Explanation:
For no reading galvanometer,\({V_{AB}} = {V_{CD}}\). \(\left( {\frac{{{E_1} - {E_2}}}{{400}}} \right) = \frac{{{E_2}}}{R}\) \(\left( {\frac{{10 - 2}}{{400}}} \right) = \frac{2}{R} \Rightarrow R = 100\,\Omega \).
NEET - 2023
PHXII03:CURRENT ELECTRICITY
357419
For the value of \(R\) in the circuit as shown in figure, current passing through \(4\Omega \) resistance will be zero.
1 \(4\Omega \)
2 \(2\Omega \)
3 \(3\Omega \)
4 \(1\Omega \)
Explanation:
Since, current does not flow in the \(4\Omega \) resistance, hence resistance \(4\Omega \) becomes ineffective in current. Current through resistance \(2\Omega \) is \(i = \frac{{9 - 6}}{2} = \frac{3}{2}A\) In circuit \(ABCDEFA\), \(9 - 3 - Ri - 2i = 0\) \( \Rightarrow R = 2\Omega \)
PHXII03:CURRENT ELECTRICITY
357420
Consider the following two statements: I. Kirchoff’s junction law follows from the conservation of charge. II. Kirchoff’s loop law follows from the conservation of energy. Which of the following is correct?
1 Both I and II are correct
2 I is wrong and II is correct
3 I is correct and II is wrong
4 Both I and II are wrong
Explanation:
Conceptual Question
PHXII03:CURRENT ELECTRICITY
357421
A \(100\,V\) voltmeter of internal resistance \(20\,k\,\Omega \) in series with a high resistance \(R\) is connected to \(110\,V\) line. The voltmeter reads \(5\,V\), the value of \(R\) is
1 \(210\,k\,\Omega \)
2 \(315\,k\,\Omega \)
3 \(420\,k\,\Omega \)
4 \(4440\,k\,\Omega \)
Explanation:
The circuit is as shown figure Potential difference across voltmeter \( = 5\;V\) \(\therefore\) Current in the circuit, \(i = \frac{5}{{{R_V}}} = \frac{5}{{20 \times {{10}^3}}} = 0.25 \times {10^{ - 3}}\;A\) Voltage across \(R,{V_1} = 110 - 5 = 105\;V\) Hence, \(R=\dfrac{V_{1}}{i}=\dfrac{105}{0.25 \times 10^{-3}}=420 \times 10^{3}=420 \mathrm{k} \Omega\) So, correct option is (3).
357418
If the galvanometer \(G\) does not show any deflection in the circuit shown, the value of \(R\) is given by:
1 \(50\,\Omega \)
2 \(100\,\Omega \)
3 \(400\,\Omega \)
4 \(200\,\Omega \)
Explanation:
For no reading galvanometer,\({V_{AB}} = {V_{CD}}\). \(\left( {\frac{{{E_1} - {E_2}}}{{400}}} \right) = \frac{{{E_2}}}{R}\) \(\left( {\frac{{10 - 2}}{{400}}} \right) = \frac{2}{R} \Rightarrow R = 100\,\Omega \).
NEET - 2023
PHXII03:CURRENT ELECTRICITY
357419
For the value of \(R\) in the circuit as shown in figure, current passing through \(4\Omega \) resistance will be zero.
1 \(4\Omega \)
2 \(2\Omega \)
3 \(3\Omega \)
4 \(1\Omega \)
Explanation:
Since, current does not flow in the \(4\Omega \) resistance, hence resistance \(4\Omega \) becomes ineffective in current. Current through resistance \(2\Omega \) is \(i = \frac{{9 - 6}}{2} = \frac{3}{2}A\) In circuit \(ABCDEFA\), \(9 - 3 - Ri - 2i = 0\) \( \Rightarrow R = 2\Omega \)
PHXII03:CURRENT ELECTRICITY
357420
Consider the following two statements: I. Kirchoff’s junction law follows from the conservation of charge. II. Kirchoff’s loop law follows from the conservation of energy. Which of the following is correct?
1 Both I and II are correct
2 I is wrong and II is correct
3 I is correct and II is wrong
4 Both I and II are wrong
Explanation:
Conceptual Question
PHXII03:CURRENT ELECTRICITY
357421
A \(100\,V\) voltmeter of internal resistance \(20\,k\,\Omega \) in series with a high resistance \(R\) is connected to \(110\,V\) line. The voltmeter reads \(5\,V\), the value of \(R\) is
1 \(210\,k\,\Omega \)
2 \(315\,k\,\Omega \)
3 \(420\,k\,\Omega \)
4 \(4440\,k\,\Omega \)
Explanation:
The circuit is as shown figure Potential difference across voltmeter \( = 5\;V\) \(\therefore\) Current in the circuit, \(i = \frac{5}{{{R_V}}} = \frac{5}{{20 \times {{10}^3}}} = 0.25 \times {10^{ - 3}}\;A\) Voltage across \(R,{V_1} = 110 - 5 = 105\;V\) Hence, \(R=\dfrac{V_{1}}{i}=\dfrac{105}{0.25 \times 10^{-3}}=420 \times 10^{3}=420 \mathrm{k} \Omega\) So, correct option is (3).
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
PHXII03:CURRENT ELECTRICITY
357418
If the galvanometer \(G\) does not show any deflection in the circuit shown, the value of \(R\) is given by:
1 \(50\,\Omega \)
2 \(100\,\Omega \)
3 \(400\,\Omega \)
4 \(200\,\Omega \)
Explanation:
For no reading galvanometer,\({V_{AB}} = {V_{CD}}\). \(\left( {\frac{{{E_1} - {E_2}}}{{400}}} \right) = \frac{{{E_2}}}{R}\) \(\left( {\frac{{10 - 2}}{{400}}} \right) = \frac{2}{R} \Rightarrow R = 100\,\Omega \).
NEET - 2023
PHXII03:CURRENT ELECTRICITY
357419
For the value of \(R\) in the circuit as shown in figure, current passing through \(4\Omega \) resistance will be zero.
1 \(4\Omega \)
2 \(2\Omega \)
3 \(3\Omega \)
4 \(1\Omega \)
Explanation:
Since, current does not flow in the \(4\Omega \) resistance, hence resistance \(4\Omega \) becomes ineffective in current. Current through resistance \(2\Omega \) is \(i = \frac{{9 - 6}}{2} = \frac{3}{2}A\) In circuit \(ABCDEFA\), \(9 - 3 - Ri - 2i = 0\) \( \Rightarrow R = 2\Omega \)
PHXII03:CURRENT ELECTRICITY
357420
Consider the following two statements: I. Kirchoff’s junction law follows from the conservation of charge. II. Kirchoff’s loop law follows from the conservation of energy. Which of the following is correct?
1 Both I and II are correct
2 I is wrong and II is correct
3 I is correct and II is wrong
4 Both I and II are wrong
Explanation:
Conceptual Question
PHXII03:CURRENT ELECTRICITY
357421
A \(100\,V\) voltmeter of internal resistance \(20\,k\,\Omega \) in series with a high resistance \(R\) is connected to \(110\,V\) line. The voltmeter reads \(5\,V\), the value of \(R\) is
1 \(210\,k\,\Omega \)
2 \(315\,k\,\Omega \)
3 \(420\,k\,\Omega \)
4 \(4440\,k\,\Omega \)
Explanation:
The circuit is as shown figure Potential difference across voltmeter \( = 5\;V\) \(\therefore\) Current in the circuit, \(i = \frac{5}{{{R_V}}} = \frac{5}{{20 \times {{10}^3}}} = 0.25 \times {10^{ - 3}}\;A\) Voltage across \(R,{V_1} = 110 - 5 = 105\;V\) Hence, \(R=\dfrac{V_{1}}{i}=\dfrac{105}{0.25 \times 10^{-3}}=420 \times 10^{3}=420 \mathrm{k} \Omega\) So, correct option is (3).
