Applying Kirchoff’s Voltage law (\(KVL\)) on the larger loop containing \(R\),\({E_2}\) and \({r_2}\) \( - \left( {{i_1} + {i_2}} \right)R + {E_2} - {i_2}{r_2} = 0\,\,\,\,\,(1)\) Applying in \(KVL\) on the upper loop containing \(R\), \({r_1}\) and \({E_1}\) \( - \left( {{i_1} + {i_2}} \right)R - {i_1}{r_1} + {E_1} = 0\,\,\,\,\,(2)\) Applying \(KVL\) on the lower loop containing \({E_1}\),\({r_1}\) , \({E_2}\) and \({r_2}\) \( - {E_1} + {i_1}{r_1} + {E_2} - {i_2}{r_2} = 0\,\,\,\,(3)\) From the given options, only option (3) is same as eq.(2) , hence it is the correct equation
PHXII03:CURRENT ELECTRICITY
357415
The figure below shows current in a part of electic circuit. The current \(I\) is
1 \(1.7\,A\)
2 \(13\,A\)
3 \(1.3\,A\)
4 \(1\,A\)
Explanation:
According to Krichhoff's first law, At junction \(A, I_{A B}=2+2=4 A\) At junction \(C,I = {I_{BC}} - 1.3 = 3 - 1.3 = 1.7\;A\)
PHXII03:CURRENT ELECTRICITY
357416
Three resistors \(1\, \Omega, 2\, \Omega\) 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:
The circuit is shown in figure. In the given circuit, \(1\, \Omega\) and \(2\, \Omega\) are in series and their series combination is in parallel with \(3 \,\Omega\). \(\therefore \quad\) The equivalent resistance of the circuit is \({R_{eq}} = \frac{{(1\,\Omega + 2\,\Omega )(3\,\Omega )}}{{(1\,\Omega + 2\,\Omega ) + (3\,\Omega )}} = \frac{3}{2}\,\Omega \) The current drawn from the battery is \(I = \frac{\varepsilon }{{{R_{eq}}}} = \frac{{3\;V}}{3} = 2\;A\) As \(R_{A C}(=3 \,\Omega)\) and \(R_{A B C}(=1\, \Omega+2\, \Omega=3\, \Omega)\) are equal. \(\therefore \quad\) The current through \(3\, \Omega\) resistor\( = \frac{I}{2} = \frac{{2\;A}}{2} = 1\;A\)
PHXII03:CURRENT ELECTRICITY
357417
If the switches \({S_{1}, S_{2}}\), and \({S_{3}}\) in figure are arranged such that the current through the battery is minimum, find the voltage across points \({A}\) and \({B}\)
1 \(1\,V\)
2 \(4\,V\)
3 \(7\,V\)
4 \(5\,V\)
Explanation:
Current through the battery will be minimum when all the three switches are open. The resulting circuit is shown in figure. Hence,\(V_{A B}=(0.5) V_{A B} \times \dfrac{24}{12}=1 {~V}\)
Applying Kirchoff’s Voltage law (\(KVL\)) on the larger loop containing \(R\),\({E_2}\) and \({r_2}\) \( - \left( {{i_1} + {i_2}} \right)R + {E_2} - {i_2}{r_2} = 0\,\,\,\,\,(1)\) Applying in \(KVL\) on the upper loop containing \(R\), \({r_1}\) and \({E_1}\) \( - \left( {{i_1} + {i_2}} \right)R - {i_1}{r_1} + {E_1} = 0\,\,\,\,\,(2)\) Applying \(KVL\) on the lower loop containing \({E_1}\),\({r_1}\) , \({E_2}\) and \({r_2}\) \( - {E_1} + {i_1}{r_1} + {E_2} - {i_2}{r_2} = 0\,\,\,\,(3)\) From the given options, only option (3) is same as eq.(2) , hence it is the correct equation
PHXII03:CURRENT ELECTRICITY
357415
The figure below shows current in a part of electic circuit. The current \(I\) is
1 \(1.7\,A\)
2 \(13\,A\)
3 \(1.3\,A\)
4 \(1\,A\)
Explanation:
According to Krichhoff's first law, At junction \(A, I_{A B}=2+2=4 A\) At junction \(C,I = {I_{BC}} - 1.3 = 3 - 1.3 = 1.7\;A\)
PHXII03:CURRENT ELECTRICITY
357416
Three resistors \(1\, \Omega, 2\, \Omega\) 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:
The circuit is shown in figure. In the given circuit, \(1\, \Omega\) and \(2\, \Omega\) are in series and their series combination is in parallel with \(3 \,\Omega\). \(\therefore \quad\) The equivalent resistance of the circuit is \({R_{eq}} = \frac{{(1\,\Omega + 2\,\Omega )(3\,\Omega )}}{{(1\,\Omega + 2\,\Omega ) + (3\,\Omega )}} = \frac{3}{2}\,\Omega \) The current drawn from the battery is \(I = \frac{\varepsilon }{{{R_{eq}}}} = \frac{{3\;V}}{3} = 2\;A\) As \(R_{A C}(=3 \,\Omega)\) and \(R_{A B C}(=1\, \Omega+2\, \Omega=3\, \Omega)\) are equal. \(\therefore \quad\) The current through \(3\, \Omega\) resistor\( = \frac{I}{2} = \frac{{2\;A}}{2} = 1\;A\)
PHXII03:CURRENT ELECTRICITY
357417
If the switches \({S_{1}, S_{2}}\), and \({S_{3}}\) in figure are arranged such that the current through the battery is minimum, find the voltage across points \({A}\) and \({B}\)
1 \(1\,V\)
2 \(4\,V\)
3 \(7\,V\)
4 \(5\,V\)
Explanation:
Current through the battery will be minimum when all the three switches are open. The resulting circuit is shown in figure. Hence,\(V_{A B}=(0.5) V_{A B} \times \dfrac{24}{12}=1 {~V}\)
Applying Kirchoff’s Voltage law (\(KVL\)) on the larger loop containing \(R\),\({E_2}\) and \({r_2}\) \( - \left( {{i_1} + {i_2}} \right)R + {E_2} - {i_2}{r_2} = 0\,\,\,\,\,(1)\) Applying in \(KVL\) on the upper loop containing \(R\), \({r_1}\) and \({E_1}\) \( - \left( {{i_1} + {i_2}} \right)R - {i_1}{r_1} + {E_1} = 0\,\,\,\,\,(2)\) Applying \(KVL\) on the lower loop containing \({E_1}\),\({r_1}\) , \({E_2}\) and \({r_2}\) \( - {E_1} + {i_1}{r_1} + {E_2} - {i_2}{r_2} = 0\,\,\,\,(3)\) From the given options, only option (3) is same as eq.(2) , hence it is the correct equation
PHXII03:CURRENT ELECTRICITY
357415
The figure below shows current in a part of electic circuit. The current \(I\) is
1 \(1.7\,A\)
2 \(13\,A\)
3 \(1.3\,A\)
4 \(1\,A\)
Explanation:
According to Krichhoff's first law, At junction \(A, I_{A B}=2+2=4 A\) At junction \(C,I = {I_{BC}} - 1.3 = 3 - 1.3 = 1.7\;A\)
PHXII03:CURRENT ELECTRICITY
357416
Three resistors \(1\, \Omega, 2\, \Omega\) 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:
The circuit is shown in figure. In the given circuit, \(1\, \Omega\) and \(2\, \Omega\) are in series and their series combination is in parallel with \(3 \,\Omega\). \(\therefore \quad\) The equivalent resistance of the circuit is \({R_{eq}} = \frac{{(1\,\Omega + 2\,\Omega )(3\,\Omega )}}{{(1\,\Omega + 2\,\Omega ) + (3\,\Omega )}} = \frac{3}{2}\,\Omega \) The current drawn from the battery is \(I = \frac{\varepsilon }{{{R_{eq}}}} = \frac{{3\;V}}{3} = 2\;A\) As \(R_{A C}(=3 \,\Omega)\) and \(R_{A B C}(=1\, \Omega+2\, \Omega=3\, \Omega)\) are equal. \(\therefore \quad\) The current through \(3\, \Omega\) resistor\( = \frac{I}{2} = \frac{{2\;A}}{2} = 1\;A\)
PHXII03:CURRENT ELECTRICITY
357417
If the switches \({S_{1}, S_{2}}\), and \({S_{3}}\) in figure are arranged such that the current through the battery is minimum, find the voltage across points \({A}\) and \({B}\)
1 \(1\,V\)
2 \(4\,V\)
3 \(7\,V\)
4 \(5\,V\)
Explanation:
Current through the battery will be minimum when all the three switches are open. The resulting circuit is shown in figure. Hence,\(V_{A B}=(0.5) V_{A B} \times \dfrac{24}{12}=1 {~V}\)
Applying Kirchoff’s Voltage law (\(KVL\)) on the larger loop containing \(R\),\({E_2}\) and \({r_2}\) \( - \left( {{i_1} + {i_2}} \right)R + {E_2} - {i_2}{r_2} = 0\,\,\,\,\,(1)\) Applying in \(KVL\) on the upper loop containing \(R\), \({r_1}\) and \({E_1}\) \( - \left( {{i_1} + {i_2}} \right)R - {i_1}{r_1} + {E_1} = 0\,\,\,\,\,(2)\) Applying \(KVL\) on the lower loop containing \({E_1}\),\({r_1}\) , \({E_2}\) and \({r_2}\) \( - {E_1} + {i_1}{r_1} + {E_2} - {i_2}{r_2} = 0\,\,\,\,(3)\) From the given options, only option (3) is same as eq.(2) , hence it is the correct equation
PHXII03:CURRENT ELECTRICITY
357415
The figure below shows current in a part of electic circuit. The current \(I\) is
1 \(1.7\,A\)
2 \(13\,A\)
3 \(1.3\,A\)
4 \(1\,A\)
Explanation:
According to Krichhoff's first law, At junction \(A, I_{A B}=2+2=4 A\) At junction \(C,I = {I_{BC}} - 1.3 = 3 - 1.3 = 1.7\;A\)
PHXII03:CURRENT ELECTRICITY
357416
Three resistors \(1\, \Omega, 2\, \Omega\) 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:
The circuit is shown in figure. In the given circuit, \(1\, \Omega\) and \(2\, \Omega\) are in series and their series combination is in parallel with \(3 \,\Omega\). \(\therefore \quad\) The equivalent resistance of the circuit is \({R_{eq}} = \frac{{(1\,\Omega + 2\,\Omega )(3\,\Omega )}}{{(1\,\Omega + 2\,\Omega ) + (3\,\Omega )}} = \frac{3}{2}\,\Omega \) The current drawn from the battery is \(I = \frac{\varepsilon }{{{R_{eq}}}} = \frac{{3\;V}}{3} = 2\;A\) As \(R_{A C}(=3 \,\Omega)\) and \(R_{A B C}(=1\, \Omega+2\, \Omega=3\, \Omega)\) are equal. \(\therefore \quad\) The current through \(3\, \Omega\) resistor\( = \frac{I}{2} = \frac{{2\;A}}{2} = 1\;A\)
PHXII03:CURRENT ELECTRICITY
357417
If the switches \({S_{1}, S_{2}}\), and \({S_{3}}\) in figure are arranged such that the current through the battery is minimum, find the voltage across points \({A}\) and \({B}\)
1 \(1\,V\)
2 \(4\,V\)
3 \(7\,V\)
4 \(5\,V\)
Explanation:
Current through the battery will be minimum when all the three switches are open. The resulting circuit is shown in figure. Hence,\(V_{A B}=(0.5) V_{A B} \times \dfrac{24}{12}=1 {~V}\)
Applying Kirchoff’s Voltage law (\(KVL\)) on the larger loop containing \(R\),\({E_2}\) and \({r_2}\) \( - \left( {{i_1} + {i_2}} \right)R + {E_2} - {i_2}{r_2} = 0\,\,\,\,\,(1)\) Applying in \(KVL\) on the upper loop containing \(R\), \({r_1}\) and \({E_1}\) \( - \left( {{i_1} + {i_2}} \right)R - {i_1}{r_1} + {E_1} = 0\,\,\,\,\,(2)\) Applying \(KVL\) on the lower loop containing \({E_1}\),\({r_1}\) , \({E_2}\) and \({r_2}\) \( - {E_1} + {i_1}{r_1} + {E_2} - {i_2}{r_2} = 0\,\,\,\,(3)\) From the given options, only option (3) is same as eq.(2) , hence it is the correct equation
PHXII03:CURRENT ELECTRICITY
357415
The figure below shows current in a part of electic circuit. The current \(I\) is
1 \(1.7\,A\)
2 \(13\,A\)
3 \(1.3\,A\)
4 \(1\,A\)
Explanation:
According to Krichhoff's first law, At junction \(A, I_{A B}=2+2=4 A\) At junction \(C,I = {I_{BC}} - 1.3 = 3 - 1.3 = 1.7\;A\)
PHXII03:CURRENT ELECTRICITY
357416
Three resistors \(1\, \Omega, 2\, \Omega\) 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:
The circuit is shown in figure. In the given circuit, \(1\, \Omega\) and \(2\, \Omega\) are in series and their series combination is in parallel with \(3 \,\Omega\). \(\therefore \quad\) The equivalent resistance of the circuit is \({R_{eq}} = \frac{{(1\,\Omega + 2\,\Omega )(3\,\Omega )}}{{(1\,\Omega + 2\,\Omega ) + (3\,\Omega )}} = \frac{3}{2}\,\Omega \) The current drawn from the battery is \(I = \frac{\varepsilon }{{{R_{eq}}}} = \frac{{3\;V}}{3} = 2\;A\) As \(R_{A C}(=3 \,\Omega)\) and \(R_{A B C}(=1\, \Omega+2\, \Omega=3\, \Omega)\) are equal. \(\therefore \quad\) The current through \(3\, \Omega\) resistor\( = \frac{I}{2} = \frac{{2\;A}}{2} = 1\;A\)
PHXII03:CURRENT ELECTRICITY
357417
If the switches \({S_{1}, S_{2}}\), and \({S_{3}}\) in figure are arranged such that the current through the battery is minimum, find the voltage across points \({A}\) and \({B}\)
1 \(1\,V\)
2 \(4\,V\)
3 \(7\,V\)
4 \(5\,V\)
Explanation:
Current through the battery will be minimum when all the three switches are open. The resulting circuit is shown in figure. Hence,\(V_{A B}=(0.5) V_{A B} \times \dfrac{24}{12}=1 {~V}\)
