NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXII03:CURRENT ELECTRICITY
357379
Three resistances \(2\Omega ,3\Omega \,{\rm{and}}\,4\Omega \) are connected in parallel. The ratio of currents passing through them when a potential difference is applied across its ends will be
1 \(6:4:3\)
2 \(4:3:2\)
3 \(6:3:2\)
4 \(5:4:3\)
Explanation:
When resistances are conneted in parallel,\(I \propto \frac{1}{R}\) \({I_1}:{I_2}:{I_3} = \frac{1}{{{R_1}}}:\frac{1}{{{R_2}}}:\frac{1}{{{R_3}}}\) \( = \frac{1}{2}:\frac{1}{3}:\frac{1}{4} = \frac{{12}}{2}:\frac{{12}}{3}:\frac{{12}}{4} = 6:4:3\)
KCET - 2015
PHXII03:CURRENT ELECTRICITY
357380
In the following electrical network, the value of ' \({I}\) ' is
357381
In the electric network shown, when no current flows through the \(4\Omega \) resistor in the arm \(EB\), the potential difference between the points \(A\) and \(D\) will be:
1 6 \(V\)
2 3 \(V\)
3 5 \(V\)
4 4 \(V\)
Explanation:
Let 0\(V\) is the potential at point \(A\). By moving along the open path \(ABED\). The potentials at other points are shown in the figure. \({V_A} = 0V,\,{V_B} = - 9V,\,{V_E} = - 5V\) and \({{\text{V}}_D} = - 5V\) \({V_A} - {V_D} = 5V\)
JEE - 2015
PHXII03:CURRENT ELECTRICITY
357382
A \(3\;V\) battery with negligible internal resistance is connected in a circuit as shown in the figure. The current \(I\), in the circuit will be
357379
Three resistances \(2\Omega ,3\Omega \,{\rm{and}}\,4\Omega \) are connected in parallel. The ratio of currents passing through them when a potential difference is applied across its ends will be
1 \(6:4:3\)
2 \(4:3:2\)
3 \(6:3:2\)
4 \(5:4:3\)
Explanation:
When resistances are conneted in parallel,\(I \propto \frac{1}{R}\) \({I_1}:{I_2}:{I_3} = \frac{1}{{{R_1}}}:\frac{1}{{{R_2}}}:\frac{1}{{{R_3}}}\) \( = \frac{1}{2}:\frac{1}{3}:\frac{1}{4} = \frac{{12}}{2}:\frac{{12}}{3}:\frac{{12}}{4} = 6:4:3\)
KCET - 2015
PHXII03:CURRENT ELECTRICITY
357380
In the following electrical network, the value of ' \({I}\) ' is
357381
In the electric network shown, when no current flows through the \(4\Omega \) resistor in the arm \(EB\), the potential difference between the points \(A\) and \(D\) will be:
1 6 \(V\)
2 3 \(V\)
3 5 \(V\)
4 4 \(V\)
Explanation:
Let 0\(V\) is the potential at point \(A\). By moving along the open path \(ABED\). The potentials at other points are shown in the figure. \({V_A} = 0V,\,{V_B} = - 9V,\,{V_E} = - 5V\) and \({{\text{V}}_D} = - 5V\) \({V_A} - {V_D} = 5V\)
JEE - 2015
PHXII03:CURRENT ELECTRICITY
357382
A \(3\;V\) battery with negligible internal resistance is connected in a circuit as shown in the figure. The current \(I\), in the circuit will be
357379
Three resistances \(2\Omega ,3\Omega \,{\rm{and}}\,4\Omega \) are connected in parallel. The ratio of currents passing through them when a potential difference is applied across its ends will be
1 \(6:4:3\)
2 \(4:3:2\)
3 \(6:3:2\)
4 \(5:4:3\)
Explanation:
When resistances are conneted in parallel,\(I \propto \frac{1}{R}\) \({I_1}:{I_2}:{I_3} = \frac{1}{{{R_1}}}:\frac{1}{{{R_2}}}:\frac{1}{{{R_3}}}\) \( = \frac{1}{2}:\frac{1}{3}:\frac{1}{4} = \frac{{12}}{2}:\frac{{12}}{3}:\frac{{12}}{4} = 6:4:3\)
KCET - 2015
PHXII03:CURRENT ELECTRICITY
357380
In the following electrical network, the value of ' \({I}\) ' is
357381
In the electric network shown, when no current flows through the \(4\Omega \) resistor in the arm \(EB\), the potential difference between the points \(A\) and \(D\) will be:
1 6 \(V\)
2 3 \(V\)
3 5 \(V\)
4 4 \(V\)
Explanation:
Let 0\(V\) is the potential at point \(A\). By moving along the open path \(ABED\). The potentials at other points are shown in the figure. \({V_A} = 0V,\,{V_B} = - 9V,\,{V_E} = - 5V\) and \({{\text{V}}_D} = - 5V\) \({V_A} - {V_D} = 5V\)
JEE - 2015
PHXII03:CURRENT ELECTRICITY
357382
A \(3\;V\) battery with negligible internal resistance is connected in a circuit as shown in the figure. The current \(I\), in the circuit will be
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
PHXII03:CURRENT ELECTRICITY
357379
Three resistances \(2\Omega ,3\Omega \,{\rm{and}}\,4\Omega \) are connected in parallel. The ratio of currents passing through them when a potential difference is applied across its ends will be
1 \(6:4:3\)
2 \(4:3:2\)
3 \(6:3:2\)
4 \(5:4:3\)
Explanation:
When resistances are conneted in parallel,\(I \propto \frac{1}{R}\) \({I_1}:{I_2}:{I_3} = \frac{1}{{{R_1}}}:\frac{1}{{{R_2}}}:\frac{1}{{{R_3}}}\) \( = \frac{1}{2}:\frac{1}{3}:\frac{1}{4} = \frac{{12}}{2}:\frac{{12}}{3}:\frac{{12}}{4} = 6:4:3\)
KCET - 2015
PHXII03:CURRENT ELECTRICITY
357380
In the following electrical network, the value of ' \({I}\) ' is
357381
In the electric network shown, when no current flows through the \(4\Omega \) resistor in the arm \(EB\), the potential difference between the points \(A\) and \(D\) will be:
1 6 \(V\)
2 3 \(V\)
3 5 \(V\)
4 4 \(V\)
Explanation:
Let 0\(V\) is the potential at point \(A\). By moving along the open path \(ABED\). The potentials at other points are shown in the figure. \({V_A} = 0V,\,{V_B} = - 9V,\,{V_E} = - 5V\) and \({{\text{V}}_D} = - 5V\) \({V_A} - {V_D} = 5V\)
JEE - 2015
PHXII03:CURRENT ELECTRICITY
357382
A \(3\;V\) battery with negligible internal resistance is connected in a circuit as shown in the figure. The current \(I\), in the circuit will be
