357443
In the shown circuit,\({V_A} - {V_B} = 16V\) The current passing through \(2\,\Omega \) resistor will be
1 \(4A\)
2 \(3.5A\)
3 \({\rm{Zero}}\)
4 \(2.5A\)
Explanation:
Let \(I\) and \({I_1}\) are current distributions in the circuit. By applying Kirchhoff’s law’s we get \( - 4I - 9 - (I - {I_1}) + 3 - 4I = {V_B} - {V_A}\) \(10 = 9I - {I_1}\,\,\,\,\,(1)\) \(3{I_1} - I = 9\,\,\,\,\,\,\,(2)\) From eqs. (1) and (2), \({I_1} = 3.5\,A\)
PHXII03:CURRENT ELECTRICITY
357444
In the given circuit diagram, the currents, \({I_1} = 0.3\,A,{I_4} = 0.8\,A\) and \({I_5} = 0.4\,A,\) are flowing as shown. The currents \({I_2},{I_3}\) and \({I_6},\) respectively, are
1 \(1.1A,\,0.4\,A,\,0.4\,A\)
2 \( - 0.4\,A,\,0.4\,A,\,1.1\,A\)
3 \(1.1A,\, - 0.4\,A,\,0.4\,A\)
4 \(0.4\,A,\,1.1\,A,0.4A\)
Explanation:
Applying \(KCL\) at point \(P,{I_5} = {I_6}\) \( \Rightarrow {I_6} = 0.4A\) From \(KCL\) at point \(R,{I_1} + {I_2} = {I_4}\) \( \Rightarrow {I_2} = {I_4} - {I_1} = 0.8 - \left( { - 0.3} \right) = 1.1\,A\) At point \(S,\,{I_5} + {I_3} = {I_4} \Rightarrow {I_3} = {I_4} - {I_5}\) \( \Rightarrow {I_3} = 0.8 - 0.4 = 0.4\,A.\)
JEE - 2019
PHXII03:CURRENT ELECTRICITY
357445
Assertion : The algebraic sum of currents meeting at junction in a closed circuit is zero. Reason : The Kirchoff's first law does not obey law of conservation of charge.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Kirchhoff's first law states that the algebraic sum of currents at any junction in a closed circuit is always zero, reflecting the conservation of charge. It ensures that the total charge entering a junction equals the total charge leaving, \(\Rightarrow\) the law of conservation of charge within the circuit holds good. So correct option is (3).
PHXII03:CURRENT ELECTRICITY
357446
In the given circuit, the current in resistance \(R_{3}\) is
1 \(1.5\,A\)
2 \(2\,A\)
3 \(2.5\,A\)
4 \(1\,A\)
Explanation:
\(R_{2}\) and \(R_{3}\) are in parallel, so \({R_p} = \frac{{4 \times 4}}{{4 + 4}} = 2\,\Omega \) \(R_{1}, R_{p}, R_{4}\) in series, \({R_{eq}} = 2 + 2 + 1 = 5\,\Omega \) \(i = \frac{V}{{{R_{eq}}}} = \frac{{10}}{5} = 2\;A\)
Current in \(R_{3}\) is \(\frac{i}{2} = \frac{2}{2} = 1\;A\)
JEE - 2024
PHXII03:CURRENT ELECTRICITY
357447
Find current in the branch \({C D}\) of the circuit shown in figure.
1 \(15\,A\)
2 \(10\,A\)
3 \(20\,A\)
4 \(8\,A\)
Explanation:
Circuit can be redrawn as shown in figure below
The equivalent resistance across the battery, \({R_{{eq}}=\dfrac{3}{2} \Omega}\) \({\Rightarrow \quad I=\dfrac{V}{R_{\text {eq }}}=20 {~A}}\) Current in \({I_{C D}=I_{A C}+I_{B} \Rightarrow I_{C D}=15 {~A}}\)
357443
In the shown circuit,\({V_A} - {V_B} = 16V\) The current passing through \(2\,\Omega \) resistor will be
1 \(4A\)
2 \(3.5A\)
3 \({\rm{Zero}}\)
4 \(2.5A\)
Explanation:
Let \(I\) and \({I_1}\) are current distributions in the circuit. By applying Kirchhoff’s law’s we get \( - 4I - 9 - (I - {I_1}) + 3 - 4I = {V_B} - {V_A}\) \(10 = 9I - {I_1}\,\,\,\,\,(1)\) \(3{I_1} - I = 9\,\,\,\,\,\,\,(2)\) From eqs. (1) and (2), \({I_1} = 3.5\,A\)
PHXII03:CURRENT ELECTRICITY
357444
In the given circuit diagram, the currents, \({I_1} = 0.3\,A,{I_4} = 0.8\,A\) and \({I_5} = 0.4\,A,\) are flowing as shown. The currents \({I_2},{I_3}\) and \({I_6},\) respectively, are
1 \(1.1A,\,0.4\,A,\,0.4\,A\)
2 \( - 0.4\,A,\,0.4\,A,\,1.1\,A\)
3 \(1.1A,\, - 0.4\,A,\,0.4\,A\)
4 \(0.4\,A,\,1.1\,A,0.4A\)
Explanation:
Applying \(KCL\) at point \(P,{I_5} = {I_6}\) \( \Rightarrow {I_6} = 0.4A\) From \(KCL\) at point \(R,{I_1} + {I_2} = {I_4}\) \( \Rightarrow {I_2} = {I_4} - {I_1} = 0.8 - \left( { - 0.3} \right) = 1.1\,A\) At point \(S,\,{I_5} + {I_3} = {I_4} \Rightarrow {I_3} = {I_4} - {I_5}\) \( \Rightarrow {I_3} = 0.8 - 0.4 = 0.4\,A.\)
JEE - 2019
PHXII03:CURRENT ELECTRICITY
357445
Assertion : The algebraic sum of currents meeting at junction in a closed circuit is zero. Reason : The Kirchoff's first law does not obey law of conservation of charge.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Kirchhoff's first law states that the algebraic sum of currents at any junction in a closed circuit is always zero, reflecting the conservation of charge. It ensures that the total charge entering a junction equals the total charge leaving, \(\Rightarrow\) the law of conservation of charge within the circuit holds good. So correct option is (3).
PHXII03:CURRENT ELECTRICITY
357446
In the given circuit, the current in resistance \(R_{3}\) is
1 \(1.5\,A\)
2 \(2\,A\)
3 \(2.5\,A\)
4 \(1\,A\)
Explanation:
\(R_{2}\) and \(R_{3}\) are in parallel, so \({R_p} = \frac{{4 \times 4}}{{4 + 4}} = 2\,\Omega \) \(R_{1}, R_{p}, R_{4}\) in series, \({R_{eq}} = 2 + 2 + 1 = 5\,\Omega \) \(i = \frac{V}{{{R_{eq}}}} = \frac{{10}}{5} = 2\;A\)
Current in \(R_{3}\) is \(\frac{i}{2} = \frac{2}{2} = 1\;A\)
JEE - 2024
PHXII03:CURRENT ELECTRICITY
357447
Find current in the branch \({C D}\) of the circuit shown in figure.
1 \(15\,A\)
2 \(10\,A\)
3 \(20\,A\)
4 \(8\,A\)
Explanation:
Circuit can be redrawn as shown in figure below
The equivalent resistance across the battery, \({R_{{eq}}=\dfrac{3}{2} \Omega}\) \({\Rightarrow \quad I=\dfrac{V}{R_{\text {eq }}}=20 {~A}}\) Current in \({I_{C D}=I_{A C}+I_{B} \Rightarrow I_{C D}=15 {~A}}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXII03:CURRENT ELECTRICITY
357443
In the shown circuit,\({V_A} - {V_B} = 16V\) The current passing through \(2\,\Omega \) resistor will be
1 \(4A\)
2 \(3.5A\)
3 \({\rm{Zero}}\)
4 \(2.5A\)
Explanation:
Let \(I\) and \({I_1}\) are current distributions in the circuit. By applying Kirchhoff’s law’s we get \( - 4I - 9 - (I - {I_1}) + 3 - 4I = {V_B} - {V_A}\) \(10 = 9I - {I_1}\,\,\,\,\,(1)\) \(3{I_1} - I = 9\,\,\,\,\,\,\,(2)\) From eqs. (1) and (2), \({I_1} = 3.5\,A\)
PHXII03:CURRENT ELECTRICITY
357444
In the given circuit diagram, the currents, \({I_1} = 0.3\,A,{I_4} = 0.8\,A\) and \({I_5} = 0.4\,A,\) are flowing as shown. The currents \({I_2},{I_3}\) and \({I_6},\) respectively, are
1 \(1.1A,\,0.4\,A,\,0.4\,A\)
2 \( - 0.4\,A,\,0.4\,A,\,1.1\,A\)
3 \(1.1A,\, - 0.4\,A,\,0.4\,A\)
4 \(0.4\,A,\,1.1\,A,0.4A\)
Explanation:
Applying \(KCL\) at point \(P,{I_5} = {I_6}\) \( \Rightarrow {I_6} = 0.4A\) From \(KCL\) at point \(R,{I_1} + {I_2} = {I_4}\) \( \Rightarrow {I_2} = {I_4} - {I_1} = 0.8 - \left( { - 0.3} \right) = 1.1\,A\) At point \(S,\,{I_5} + {I_3} = {I_4} \Rightarrow {I_3} = {I_4} - {I_5}\) \( \Rightarrow {I_3} = 0.8 - 0.4 = 0.4\,A.\)
JEE - 2019
PHXII03:CURRENT ELECTRICITY
357445
Assertion : The algebraic sum of currents meeting at junction in a closed circuit is zero. Reason : The Kirchoff's first law does not obey law of conservation of charge.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Kirchhoff's first law states that the algebraic sum of currents at any junction in a closed circuit is always zero, reflecting the conservation of charge. It ensures that the total charge entering a junction equals the total charge leaving, \(\Rightarrow\) the law of conservation of charge within the circuit holds good. So correct option is (3).
PHXII03:CURRENT ELECTRICITY
357446
In the given circuit, the current in resistance \(R_{3}\) is
1 \(1.5\,A\)
2 \(2\,A\)
3 \(2.5\,A\)
4 \(1\,A\)
Explanation:
\(R_{2}\) and \(R_{3}\) are in parallel, so \({R_p} = \frac{{4 \times 4}}{{4 + 4}} = 2\,\Omega \) \(R_{1}, R_{p}, R_{4}\) in series, \({R_{eq}} = 2 + 2 + 1 = 5\,\Omega \) \(i = \frac{V}{{{R_{eq}}}} = \frac{{10}}{5} = 2\;A\)
Current in \(R_{3}\) is \(\frac{i}{2} = \frac{2}{2} = 1\;A\)
JEE - 2024
PHXII03:CURRENT ELECTRICITY
357447
Find current in the branch \({C D}\) of the circuit shown in figure.
1 \(15\,A\)
2 \(10\,A\)
3 \(20\,A\)
4 \(8\,A\)
Explanation:
Circuit can be redrawn as shown in figure below
The equivalent resistance across the battery, \({R_{{eq}}=\dfrac{3}{2} \Omega}\) \({\Rightarrow \quad I=\dfrac{V}{R_{\text {eq }}}=20 {~A}}\) Current in \({I_{C D}=I_{A C}+I_{B} \Rightarrow I_{C D}=15 {~A}}\)
357443
In the shown circuit,\({V_A} - {V_B} = 16V\) The current passing through \(2\,\Omega \) resistor will be
1 \(4A\)
2 \(3.5A\)
3 \({\rm{Zero}}\)
4 \(2.5A\)
Explanation:
Let \(I\) and \({I_1}\) are current distributions in the circuit. By applying Kirchhoff’s law’s we get \( - 4I - 9 - (I - {I_1}) + 3 - 4I = {V_B} - {V_A}\) \(10 = 9I - {I_1}\,\,\,\,\,(1)\) \(3{I_1} - I = 9\,\,\,\,\,\,\,(2)\) From eqs. (1) and (2), \({I_1} = 3.5\,A\)
PHXII03:CURRENT ELECTRICITY
357444
In the given circuit diagram, the currents, \({I_1} = 0.3\,A,{I_4} = 0.8\,A\) and \({I_5} = 0.4\,A,\) are flowing as shown. The currents \({I_2},{I_3}\) and \({I_6},\) respectively, are
1 \(1.1A,\,0.4\,A,\,0.4\,A\)
2 \( - 0.4\,A,\,0.4\,A,\,1.1\,A\)
3 \(1.1A,\, - 0.4\,A,\,0.4\,A\)
4 \(0.4\,A,\,1.1\,A,0.4A\)
Explanation:
Applying \(KCL\) at point \(P,{I_5} = {I_6}\) \( \Rightarrow {I_6} = 0.4A\) From \(KCL\) at point \(R,{I_1} + {I_2} = {I_4}\) \( \Rightarrow {I_2} = {I_4} - {I_1} = 0.8 - \left( { - 0.3} \right) = 1.1\,A\) At point \(S,\,{I_5} + {I_3} = {I_4} \Rightarrow {I_3} = {I_4} - {I_5}\) \( \Rightarrow {I_3} = 0.8 - 0.4 = 0.4\,A.\)
JEE - 2019
PHXII03:CURRENT ELECTRICITY
357445
Assertion : The algebraic sum of currents meeting at junction in a closed circuit is zero. Reason : The Kirchoff's first law does not obey law of conservation of charge.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Kirchhoff's first law states that the algebraic sum of currents at any junction in a closed circuit is always zero, reflecting the conservation of charge. It ensures that the total charge entering a junction equals the total charge leaving, \(\Rightarrow\) the law of conservation of charge within the circuit holds good. So correct option is (3).
PHXII03:CURRENT ELECTRICITY
357446
In the given circuit, the current in resistance \(R_{3}\) is
1 \(1.5\,A\)
2 \(2\,A\)
3 \(2.5\,A\)
4 \(1\,A\)
Explanation:
\(R_{2}\) and \(R_{3}\) are in parallel, so \({R_p} = \frac{{4 \times 4}}{{4 + 4}} = 2\,\Omega \) \(R_{1}, R_{p}, R_{4}\) in series, \({R_{eq}} = 2 + 2 + 1 = 5\,\Omega \) \(i = \frac{V}{{{R_{eq}}}} = \frac{{10}}{5} = 2\;A\)
Current in \(R_{3}\) is \(\frac{i}{2} = \frac{2}{2} = 1\;A\)
JEE - 2024
PHXII03:CURRENT ELECTRICITY
357447
Find current in the branch \({C D}\) of the circuit shown in figure.
1 \(15\,A\)
2 \(10\,A\)
3 \(20\,A\)
4 \(8\,A\)
Explanation:
Circuit can be redrawn as shown in figure below
The equivalent resistance across the battery, \({R_{{eq}}=\dfrac{3}{2} \Omega}\) \({\Rightarrow \quad I=\dfrac{V}{R_{\text {eq }}}=20 {~A}}\) Current in \({I_{C D}=I_{A C}+I_{B} \Rightarrow I_{C D}=15 {~A}}\)
357443
In the shown circuit,\({V_A} - {V_B} = 16V\) The current passing through \(2\,\Omega \) resistor will be
1 \(4A\)
2 \(3.5A\)
3 \({\rm{Zero}}\)
4 \(2.5A\)
Explanation:
Let \(I\) and \({I_1}\) are current distributions in the circuit. By applying Kirchhoff’s law’s we get \( - 4I - 9 - (I - {I_1}) + 3 - 4I = {V_B} - {V_A}\) \(10 = 9I - {I_1}\,\,\,\,\,(1)\) \(3{I_1} - I = 9\,\,\,\,\,\,\,(2)\) From eqs. (1) and (2), \({I_1} = 3.5\,A\)
PHXII03:CURRENT ELECTRICITY
357444
In the given circuit diagram, the currents, \({I_1} = 0.3\,A,{I_4} = 0.8\,A\) and \({I_5} = 0.4\,A,\) are flowing as shown. The currents \({I_2},{I_3}\) and \({I_6},\) respectively, are
1 \(1.1A,\,0.4\,A,\,0.4\,A\)
2 \( - 0.4\,A,\,0.4\,A,\,1.1\,A\)
3 \(1.1A,\, - 0.4\,A,\,0.4\,A\)
4 \(0.4\,A,\,1.1\,A,0.4A\)
Explanation:
Applying \(KCL\) at point \(P,{I_5} = {I_6}\) \( \Rightarrow {I_6} = 0.4A\) From \(KCL\) at point \(R,{I_1} + {I_2} = {I_4}\) \( \Rightarrow {I_2} = {I_4} - {I_1} = 0.8 - \left( { - 0.3} \right) = 1.1\,A\) At point \(S,\,{I_5} + {I_3} = {I_4} \Rightarrow {I_3} = {I_4} - {I_5}\) \( \Rightarrow {I_3} = 0.8 - 0.4 = 0.4\,A.\)
JEE - 2019
PHXII03:CURRENT ELECTRICITY
357445
Assertion : The algebraic sum of currents meeting at junction in a closed circuit is zero. Reason : The Kirchoff's first law does not obey law of conservation of charge.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Kirchhoff's first law states that the algebraic sum of currents at any junction in a closed circuit is always zero, reflecting the conservation of charge. It ensures that the total charge entering a junction equals the total charge leaving, \(\Rightarrow\) the law of conservation of charge within the circuit holds good. So correct option is (3).
PHXII03:CURRENT ELECTRICITY
357446
In the given circuit, the current in resistance \(R_{3}\) is
1 \(1.5\,A\)
2 \(2\,A\)
3 \(2.5\,A\)
4 \(1\,A\)
Explanation:
\(R_{2}\) and \(R_{3}\) are in parallel, so \({R_p} = \frac{{4 \times 4}}{{4 + 4}} = 2\,\Omega \) \(R_{1}, R_{p}, R_{4}\) in series, \({R_{eq}} = 2 + 2 + 1 = 5\,\Omega \) \(i = \frac{V}{{{R_{eq}}}} = \frac{{10}}{5} = 2\;A\)
Current in \(R_{3}\) is \(\frac{i}{2} = \frac{2}{2} = 1\;A\)
JEE - 2024
PHXII03:CURRENT ELECTRICITY
357447
Find current in the branch \({C D}\) of the circuit shown in figure.
1 \(15\,A\)
2 \(10\,A\)
3 \(20\,A\)
4 \(8\,A\)
Explanation:
Circuit can be redrawn as shown in figure below
The equivalent resistance across the battery, \({R_{{eq}}=\dfrac{3}{2} \Omega}\) \({\Rightarrow \quad I=\dfrac{V}{R_{\text {eq }}}=20 {~A}}\) Current in \({I_{C D}=I_{A C}+I_{B} \Rightarrow I_{C D}=15 {~A}}\)
