357409
In the circuit shown below, the current in the \(1\,\Omega \) resistor is
1 0.13 \(A\) from \(P\) to \(Q\)
2 0.13 \(A\) from \(Q\) to \(P\)
3 1.3 \(A\) from \(P\) to \(Q\)
4 0 \(A\)
Explanation:
Applying Kirchhoff’s loop law in loops 1 and 2 in the directions shown in figure we have \(6 - 3({i_1} + {i_2}) - {i_2} = 0\,\,\,\,\,(1)\) \(9 - 2{i_1} + {i_2} - 3{i_1} = 0\,\,\,\,\,\,(2)\) Solving eqs, (1) and (2) we get, \({i_2} = 0.13A\) Hence, the currect in \(1\,\Omega \) resistor is 0.13 \(A\) from \(Q\) to \(P\).
PHXII03:CURRENT ELECTRICITY
357410
The figure shows a network of currents. The current \(i\) will be
1 \(3\,A\)
2 \(13\,A\)
3 \(10\,A\)
4 \(8\,A\)
Explanation:
The distribution of currents is shown in the figure. From the figure \(i = 6 + 4 = 10\;A\).
PHXII03:CURRENT ELECTRICITY
357411
Three resistors \(1\Omega ,2\Omega \,{\rm{and}}\,3\Omega \) are connected to form a triangle. Across \(3\Omega \) resistor a 3\(V\) battery is connected. The current through \(3\Omega \) resistor is
1 \(0.75\,A\)
2 \(1\,A\)
3 \(2\,A\)
4 \(1.5\,A\)
Explanation:
Potential difference across \(3\Omega \) resistor \( = 3V\). \(\therefore \) Current through \(3\,\Omega \) resistor \( = {I_1} = (3V{\rm{/}}3\Omega ) = 1A\)
KCET - 2009
PHXII03:CURRENT ELECTRICITY
357412
Which of the following is the correct equation when kirchhoff’s loop rule is applied to the loop \(BCDEB\) in clockwise direction?
357409
In the circuit shown below, the current in the \(1\,\Omega \) resistor is
1 0.13 \(A\) from \(P\) to \(Q\)
2 0.13 \(A\) from \(Q\) to \(P\)
3 1.3 \(A\) from \(P\) to \(Q\)
4 0 \(A\)
Explanation:
Applying Kirchhoff’s loop law in loops 1 and 2 in the directions shown in figure we have \(6 - 3({i_1} + {i_2}) - {i_2} = 0\,\,\,\,\,(1)\) \(9 - 2{i_1} + {i_2} - 3{i_1} = 0\,\,\,\,\,\,(2)\) Solving eqs, (1) and (2) we get, \({i_2} = 0.13A\) Hence, the currect in \(1\,\Omega \) resistor is 0.13 \(A\) from \(Q\) to \(P\).
PHXII03:CURRENT ELECTRICITY
357410
The figure shows a network of currents. The current \(i\) will be
1 \(3\,A\)
2 \(13\,A\)
3 \(10\,A\)
4 \(8\,A\)
Explanation:
The distribution of currents is shown in the figure. From the figure \(i = 6 + 4 = 10\;A\).
PHXII03:CURRENT ELECTRICITY
357411
Three resistors \(1\Omega ,2\Omega \,{\rm{and}}\,3\Omega \) are connected to form a triangle. Across \(3\Omega \) resistor a 3\(V\) battery is connected. The current through \(3\Omega \) resistor is
1 \(0.75\,A\)
2 \(1\,A\)
3 \(2\,A\)
4 \(1.5\,A\)
Explanation:
Potential difference across \(3\Omega \) resistor \( = 3V\). \(\therefore \) Current through \(3\,\Omega \) resistor \( = {I_1} = (3V{\rm{/}}3\Omega ) = 1A\)
KCET - 2009
PHXII03:CURRENT ELECTRICITY
357412
Which of the following is the correct equation when kirchhoff’s loop rule is applied to the loop \(BCDEB\) in clockwise direction?
357409
In the circuit shown below, the current in the \(1\,\Omega \) resistor is
1 0.13 \(A\) from \(P\) to \(Q\)
2 0.13 \(A\) from \(Q\) to \(P\)
3 1.3 \(A\) from \(P\) to \(Q\)
4 0 \(A\)
Explanation:
Applying Kirchhoff’s loop law in loops 1 and 2 in the directions shown in figure we have \(6 - 3({i_1} + {i_2}) - {i_2} = 0\,\,\,\,\,(1)\) \(9 - 2{i_1} + {i_2} - 3{i_1} = 0\,\,\,\,\,\,(2)\) Solving eqs, (1) and (2) we get, \({i_2} = 0.13A\) Hence, the currect in \(1\,\Omega \) resistor is 0.13 \(A\) from \(Q\) to \(P\).
PHXII03:CURRENT ELECTRICITY
357410
The figure shows a network of currents. The current \(i\) will be
1 \(3\,A\)
2 \(13\,A\)
3 \(10\,A\)
4 \(8\,A\)
Explanation:
The distribution of currents is shown in the figure. From the figure \(i = 6 + 4 = 10\;A\).
PHXII03:CURRENT ELECTRICITY
357411
Three resistors \(1\Omega ,2\Omega \,{\rm{and}}\,3\Omega \) are connected to form a triangle. Across \(3\Omega \) resistor a 3\(V\) battery is connected. The current through \(3\Omega \) resistor is
1 \(0.75\,A\)
2 \(1\,A\)
3 \(2\,A\)
4 \(1.5\,A\)
Explanation:
Potential difference across \(3\Omega \) resistor \( = 3V\). \(\therefore \) Current through \(3\,\Omega \) resistor \( = {I_1} = (3V{\rm{/}}3\Omega ) = 1A\)
KCET - 2009
PHXII03:CURRENT ELECTRICITY
357412
Which of the following is the correct equation when kirchhoff’s loop rule is applied to the loop \(BCDEB\) in clockwise direction?
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
PHXII03:CURRENT ELECTRICITY
357409
In the circuit shown below, the current in the \(1\,\Omega \) resistor is
1 0.13 \(A\) from \(P\) to \(Q\)
2 0.13 \(A\) from \(Q\) to \(P\)
3 1.3 \(A\) from \(P\) to \(Q\)
4 0 \(A\)
Explanation:
Applying Kirchhoff’s loop law in loops 1 and 2 in the directions shown in figure we have \(6 - 3({i_1} + {i_2}) - {i_2} = 0\,\,\,\,\,(1)\) \(9 - 2{i_1} + {i_2} - 3{i_1} = 0\,\,\,\,\,\,(2)\) Solving eqs, (1) and (2) we get, \({i_2} = 0.13A\) Hence, the currect in \(1\,\Omega \) resistor is 0.13 \(A\) from \(Q\) to \(P\).
PHXII03:CURRENT ELECTRICITY
357410
The figure shows a network of currents. The current \(i\) will be
1 \(3\,A\)
2 \(13\,A\)
3 \(10\,A\)
4 \(8\,A\)
Explanation:
The distribution of currents is shown in the figure. From the figure \(i = 6 + 4 = 10\;A\).
PHXII03:CURRENT ELECTRICITY
357411
Three resistors \(1\Omega ,2\Omega \,{\rm{and}}\,3\Omega \) are connected to form a triangle. Across \(3\Omega \) resistor a 3\(V\) battery is connected. The current through \(3\Omega \) resistor is
1 \(0.75\,A\)
2 \(1\,A\)
3 \(2\,A\)
4 \(1.5\,A\)
Explanation:
Potential difference across \(3\Omega \) resistor \( = 3V\). \(\therefore \) Current through \(3\,\Omega \) resistor \( = {I_1} = (3V{\rm{/}}3\Omega ) = 1A\)
KCET - 2009
PHXII03:CURRENT ELECTRICITY
357412
Which of the following is the correct equation when kirchhoff’s loop rule is applied to the loop \(BCDEB\) in clockwise direction?
