357430
In the circuit shown here, \(E_{1}=E_{2}=E_{3}=2 \mathrm{~V}\) and \(R_{1}=R_{2}=4\) ohms. The current flowing between points \(A\) and \(B\) through battery \(E_{2}\) is:
1 Zero
2 2 amp from \(A\) to \(B\)
3 2 amp from \(B\) to \(A\)
4 None of these
Explanation:
Circuit can be redrawn as The upper and lower branches can be replaced with a battery of \(emf\) \(\varepsilon_{0}\) and internal resistance \(r_{0}\), which are given by \({\varepsilon _0} = \frac{{\frac{{{\varepsilon _1}}}{{{r_1}}} + \frac{{{\varepsilon _2}}}{{{r_2}}}}}{{\frac{1}{{{r_1}}} + \frac{1}{{{r_2}}}}} = \frac{{\frac{2}{4} + \frac{2}{4}}}{{\frac{1}{4} + \frac{1}{4}}} = 2\;V\) \(\dfrac{1}{r_{0}}=\dfrac{1}{r_{1}}+\dfrac{1}{r_{2}}=\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{2}{4}=\dfrac{1}{2}\) \(\Rightarrow r_{0}=2 \Omega\) Total \(emf\) \( = 2 + 2 = 4\;V\) so \(\quad I=\dfrac{4}{2}=2 A\)
PHXII03:CURRENT ELECTRICITY
357431
For the given electrical arrangement, what is the value of current \(I\) ?
1 \(6\,\;A\)
2 \(5\;\,A\)
3 \(7\;\,A\)
4 \(8\;\,A\)
Explanation:
According to Kirchhoff's Ist law, Current at \(A = 3 - 2 = 1\;A\) Current at \(B = 5 + 1 = 6\;A\) Current at \(D = 4 + 1 = 5\;A\) Current at \(C = 6 + 5 = 11\;A\) So, \(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,I = 11 - 3 = 8\;{\rm{A}}\)
PHXII03:CURRENT ELECTRICITY
357432
The current flowing through \({R_2}\) is
1 \(1 / 4 A\)
2 \(2 / 3 A\)
3 \(1 / 3 A\)
4 \(1 / 2 A\)
Explanation:
Considering the circuit from side \(A B\), we will get by series parallel combination, \(R_{c q}=2 \Omega\) Total current, \(I = \frac{V}{R} = \frac{8}{2} = 4\;A\) Current through \({R_5},I = \frac{I}{2} = \frac{4}{2};I = 2\;A\) Current through \({R_4},{I_1} = \frac{{2 \times 3}}{{3 + 6}} = \frac{2}{3}\;A\) Current through \({R_2},{I_2} = \frac{{{I_1}}}{2} = \frac{1}{3}\;A\)
JEE - 2023
PHXII03:CURRENT ELECTRICITY
357433
In the following network potential at \(O\)
357430
In the circuit shown here, \(E_{1}=E_{2}=E_{3}=2 \mathrm{~V}\) and \(R_{1}=R_{2}=4\) ohms. The current flowing between points \(A\) and \(B\) through battery \(E_{2}\) is:
1 Zero
2 2 amp from \(A\) to \(B\)
3 2 amp from \(B\) to \(A\)
4 None of these
Explanation:
Circuit can be redrawn as The upper and lower branches can be replaced with a battery of \(emf\) \(\varepsilon_{0}\) and internal resistance \(r_{0}\), which are given by \({\varepsilon _0} = \frac{{\frac{{{\varepsilon _1}}}{{{r_1}}} + \frac{{{\varepsilon _2}}}{{{r_2}}}}}{{\frac{1}{{{r_1}}} + \frac{1}{{{r_2}}}}} = \frac{{\frac{2}{4} + \frac{2}{4}}}{{\frac{1}{4} + \frac{1}{4}}} = 2\;V\) \(\dfrac{1}{r_{0}}=\dfrac{1}{r_{1}}+\dfrac{1}{r_{2}}=\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{2}{4}=\dfrac{1}{2}\) \(\Rightarrow r_{0}=2 \Omega\) Total \(emf\) \( = 2 + 2 = 4\;V\) so \(\quad I=\dfrac{4}{2}=2 A\)
PHXII03:CURRENT ELECTRICITY
357431
For the given electrical arrangement, what is the value of current \(I\) ?
1 \(6\,\;A\)
2 \(5\;\,A\)
3 \(7\;\,A\)
4 \(8\;\,A\)
Explanation:
According to Kirchhoff's Ist law, Current at \(A = 3 - 2 = 1\;A\) Current at \(B = 5 + 1 = 6\;A\) Current at \(D = 4 + 1 = 5\;A\) Current at \(C = 6 + 5 = 11\;A\) So, \(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,I = 11 - 3 = 8\;{\rm{A}}\)
PHXII03:CURRENT ELECTRICITY
357432
The current flowing through \({R_2}\) is
1 \(1 / 4 A\)
2 \(2 / 3 A\)
3 \(1 / 3 A\)
4 \(1 / 2 A\)
Explanation:
Considering the circuit from side \(A B\), we will get by series parallel combination, \(R_{c q}=2 \Omega\) Total current, \(I = \frac{V}{R} = \frac{8}{2} = 4\;A\) Current through \({R_5},I = \frac{I}{2} = \frac{4}{2};I = 2\;A\) Current through \({R_4},{I_1} = \frac{{2 \times 3}}{{3 + 6}} = \frac{2}{3}\;A\) Current through \({R_2},{I_2} = \frac{{{I_1}}}{2} = \frac{1}{3}\;A\)
JEE - 2023
PHXII03:CURRENT ELECTRICITY
357433
In the following network potential at \(O\)
357430
In the circuit shown here, \(E_{1}=E_{2}=E_{3}=2 \mathrm{~V}\) and \(R_{1}=R_{2}=4\) ohms. The current flowing between points \(A\) and \(B\) through battery \(E_{2}\) is:
1 Zero
2 2 amp from \(A\) to \(B\)
3 2 amp from \(B\) to \(A\)
4 None of these
Explanation:
Circuit can be redrawn as The upper and lower branches can be replaced with a battery of \(emf\) \(\varepsilon_{0}\) and internal resistance \(r_{0}\), which are given by \({\varepsilon _0} = \frac{{\frac{{{\varepsilon _1}}}{{{r_1}}} + \frac{{{\varepsilon _2}}}{{{r_2}}}}}{{\frac{1}{{{r_1}}} + \frac{1}{{{r_2}}}}} = \frac{{\frac{2}{4} + \frac{2}{4}}}{{\frac{1}{4} + \frac{1}{4}}} = 2\;V\) \(\dfrac{1}{r_{0}}=\dfrac{1}{r_{1}}+\dfrac{1}{r_{2}}=\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{2}{4}=\dfrac{1}{2}\) \(\Rightarrow r_{0}=2 \Omega\) Total \(emf\) \( = 2 + 2 = 4\;V\) so \(\quad I=\dfrac{4}{2}=2 A\)
PHXII03:CURRENT ELECTRICITY
357431
For the given electrical arrangement, what is the value of current \(I\) ?
1 \(6\,\;A\)
2 \(5\;\,A\)
3 \(7\;\,A\)
4 \(8\;\,A\)
Explanation:
According to Kirchhoff's Ist law, Current at \(A = 3 - 2 = 1\;A\) Current at \(B = 5 + 1 = 6\;A\) Current at \(D = 4 + 1 = 5\;A\) Current at \(C = 6 + 5 = 11\;A\) So, \(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,I = 11 - 3 = 8\;{\rm{A}}\)
PHXII03:CURRENT ELECTRICITY
357432
The current flowing through \({R_2}\) is
1 \(1 / 4 A\)
2 \(2 / 3 A\)
3 \(1 / 3 A\)
4 \(1 / 2 A\)
Explanation:
Considering the circuit from side \(A B\), we will get by series parallel combination, \(R_{c q}=2 \Omega\) Total current, \(I = \frac{V}{R} = \frac{8}{2} = 4\;A\) Current through \({R_5},I = \frac{I}{2} = \frac{4}{2};I = 2\;A\) Current through \({R_4},{I_1} = \frac{{2 \times 3}}{{3 + 6}} = \frac{2}{3}\;A\) Current through \({R_2},{I_2} = \frac{{{I_1}}}{2} = \frac{1}{3}\;A\)
JEE - 2023
PHXII03:CURRENT ELECTRICITY
357433
In the following network potential at \(O\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXII03:CURRENT ELECTRICITY
357430
In the circuit shown here, \(E_{1}=E_{2}=E_{3}=2 \mathrm{~V}\) and \(R_{1}=R_{2}=4\) ohms. The current flowing between points \(A\) and \(B\) through battery \(E_{2}\) is:
1 Zero
2 2 amp from \(A\) to \(B\)
3 2 amp from \(B\) to \(A\)
4 None of these
Explanation:
Circuit can be redrawn as The upper and lower branches can be replaced with a battery of \(emf\) \(\varepsilon_{0}\) and internal resistance \(r_{0}\), which are given by \({\varepsilon _0} = \frac{{\frac{{{\varepsilon _1}}}{{{r_1}}} + \frac{{{\varepsilon _2}}}{{{r_2}}}}}{{\frac{1}{{{r_1}}} + \frac{1}{{{r_2}}}}} = \frac{{\frac{2}{4} + \frac{2}{4}}}{{\frac{1}{4} + \frac{1}{4}}} = 2\;V\) \(\dfrac{1}{r_{0}}=\dfrac{1}{r_{1}}+\dfrac{1}{r_{2}}=\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{2}{4}=\dfrac{1}{2}\) \(\Rightarrow r_{0}=2 \Omega\) Total \(emf\) \( = 2 + 2 = 4\;V\) so \(\quad I=\dfrac{4}{2}=2 A\)
PHXII03:CURRENT ELECTRICITY
357431
For the given electrical arrangement, what is the value of current \(I\) ?
1 \(6\,\;A\)
2 \(5\;\,A\)
3 \(7\;\,A\)
4 \(8\;\,A\)
Explanation:
According to Kirchhoff's Ist law, Current at \(A = 3 - 2 = 1\;A\) Current at \(B = 5 + 1 = 6\;A\) Current at \(D = 4 + 1 = 5\;A\) Current at \(C = 6 + 5 = 11\;A\) So, \(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,I = 11 - 3 = 8\;{\rm{A}}\)
PHXII03:CURRENT ELECTRICITY
357432
The current flowing through \({R_2}\) is
1 \(1 / 4 A\)
2 \(2 / 3 A\)
3 \(1 / 3 A\)
4 \(1 / 2 A\)
Explanation:
Considering the circuit from side \(A B\), we will get by series parallel combination, \(R_{c q}=2 \Omega\) Total current, \(I = \frac{V}{R} = \frac{8}{2} = 4\;A\) Current through \({R_5},I = \frac{I}{2} = \frac{4}{2};I = 2\;A\) Current through \({R_4},{I_1} = \frac{{2 \times 3}}{{3 + 6}} = \frac{2}{3}\;A\) Current through \({R_2},{I_2} = \frac{{{I_1}}}{2} = \frac{1}{3}\;A\)
JEE - 2023
PHXII03:CURRENT ELECTRICITY
357433
In the following network potential at \(O\)
357430
In the circuit shown here, \(E_{1}=E_{2}=E_{3}=2 \mathrm{~V}\) and \(R_{1}=R_{2}=4\) ohms. The current flowing between points \(A\) and \(B\) through battery \(E_{2}\) is:
1 Zero
2 2 amp from \(A\) to \(B\)
3 2 amp from \(B\) to \(A\)
4 None of these
Explanation:
Circuit can be redrawn as The upper and lower branches can be replaced with a battery of \(emf\) \(\varepsilon_{0}\) and internal resistance \(r_{0}\), which are given by \({\varepsilon _0} = \frac{{\frac{{{\varepsilon _1}}}{{{r_1}}} + \frac{{{\varepsilon _2}}}{{{r_2}}}}}{{\frac{1}{{{r_1}}} + \frac{1}{{{r_2}}}}} = \frac{{\frac{2}{4} + \frac{2}{4}}}{{\frac{1}{4} + \frac{1}{4}}} = 2\;V\) \(\dfrac{1}{r_{0}}=\dfrac{1}{r_{1}}+\dfrac{1}{r_{2}}=\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{2}{4}=\dfrac{1}{2}\) \(\Rightarrow r_{0}=2 \Omega\) Total \(emf\) \( = 2 + 2 = 4\;V\) so \(\quad I=\dfrac{4}{2}=2 A\)
PHXII03:CURRENT ELECTRICITY
357431
For the given electrical arrangement, what is the value of current \(I\) ?
1 \(6\,\;A\)
2 \(5\;\,A\)
3 \(7\;\,A\)
4 \(8\;\,A\)
Explanation:
According to Kirchhoff's Ist law, Current at \(A = 3 - 2 = 1\;A\) Current at \(B = 5 + 1 = 6\;A\) Current at \(D = 4 + 1 = 5\;A\) Current at \(C = 6 + 5 = 11\;A\) So, \(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,I = 11 - 3 = 8\;{\rm{A}}\)
PHXII03:CURRENT ELECTRICITY
357432
The current flowing through \({R_2}\) is
1 \(1 / 4 A\)
2 \(2 / 3 A\)
3 \(1 / 3 A\)
4 \(1 / 2 A\)
Explanation:
Considering the circuit from side \(A B\), we will get by series parallel combination, \(R_{c q}=2 \Omega\) Total current, \(I = \frac{V}{R} = \frac{8}{2} = 4\;A\) Current through \({R_5},I = \frac{I}{2} = \frac{4}{2};I = 2\;A\) Current through \({R_4},{I_1} = \frac{{2 \times 3}}{{3 + 6}} = \frac{2}{3}\;A\) Current through \({R_2},{I_2} = \frac{{{I_1}}}{2} = \frac{1}{3}\;A\)
JEE - 2023
PHXII03:CURRENT ELECTRICITY
357433
In the following network potential at \(O\)
