Cell
PHXII03:CURRENT ELECTRICITY

356884 Five cells each of emf \(E\) and internal resistance \(r\) send the same amount of current through an external resistance \(R\) whether the cells are connected in parallel or in series. Then the ratio \(\left( {\frac{R}{r}} \right)\)is

1 \(\frac{1}{2}\)
2 \(2\)
3 \(1\)
4 \(\frac{1}{5}\)
PHXII03:CURRENT ELECTRICITY

356885 Two cells of internal resistances \({r_{1\,}}\,{\rm{and}}\,{r_2}\) and of same emf are connected in series across a resistor of resistance \(R\). If the terminal potential difference across the cell of internal resistance \({r_{1\,}}\) is zero, then the value of \(R\) is

1 \(R = 2({r_1} + {r_2})\)
2 \(R = {r_2} - {r_1}\)
3 \(R = {r_1} - {r_2}\)
4 \(R = 2({r_1} - {r_2})\)
PHXII03:CURRENT ELECTRICITY

356886 Ten identical cells each emf \(2 V\) and internal resistances \(1 \Omega\) are connected in series with two cells wrongly connected. A resistor of \(10 \Omega\) is connected to the combination. What is the current through the resistor?

1 \(0.6\;A\)
2 \(1.2\;A\)
3 \(1.8\;A\)
4 \(2.4\;A\)
PHXII03:CURRENT ELECTRICITY

356887 While connecting 5 cells in a battery in series, in a tape recorder, by mistake one cell is connected with reverse polarity. If the effective load resistance is \(7\,\Omega \) of load is \(24\,ohm\) and internal resistance of each cell is one ohm and emf \(2\;V\), the current delivered by the battery is

1 \(0.1\,A\)
2 \(0.5\,A\)
3 \(0.3\,A\)
4 \(0.4\,A\)
PHXII03:CURRENT ELECTRICITY

356884 Five cells each of emf \(E\) and internal resistance \(r\) send the same amount of current through an external resistance \(R\) whether the cells are connected in parallel or in series. Then the ratio \(\left( {\frac{R}{r}} \right)\)is

1 \(\frac{1}{2}\)
2 \(2\)
3 \(1\)
4 \(\frac{1}{5}\)
PHXII03:CURRENT ELECTRICITY

356885 Two cells of internal resistances \({r_{1\,}}\,{\rm{and}}\,{r_2}\) and of same emf are connected in series across a resistor of resistance \(R\). If the terminal potential difference across the cell of internal resistance \({r_{1\,}}\) is zero, then the value of \(R\) is

1 \(R = 2({r_1} + {r_2})\)
2 \(R = {r_2} - {r_1}\)
3 \(R = {r_1} - {r_2}\)
4 \(R = 2({r_1} - {r_2})\)
PHXII03:CURRENT ELECTRICITY

356886 Ten identical cells each emf \(2 V\) and internal resistances \(1 \Omega\) are connected in series with two cells wrongly connected. A resistor of \(10 \Omega\) is connected to the combination. What is the current through the resistor?

1 \(0.6\;A\)
2 \(1.2\;A\)
3 \(1.8\;A\)
4 \(2.4\;A\)
PHXII03:CURRENT ELECTRICITY

356887 While connecting 5 cells in a battery in series, in a tape recorder, by mistake one cell is connected with reverse polarity. If the effective load resistance is \(7\,\Omega \) of load is \(24\,ohm\) and internal resistance of each cell is one ohm and emf \(2\;V\), the current delivered by the battery is

1 \(0.1\,A\)
2 \(0.5\,A\)
3 \(0.3\,A\)
4 \(0.4\,A\)
NEET Test Series from KOTA - 10 Papers In MS WORD WhatsApp Here
PHXII03:CURRENT ELECTRICITY

356884 Five cells each of emf \(E\) and internal resistance \(r\) send the same amount of current through an external resistance \(R\) whether the cells are connected in parallel or in series. Then the ratio \(\left( {\frac{R}{r}} \right)\)is

1 \(\frac{1}{2}\)
2 \(2\)
3 \(1\)
4 \(\frac{1}{5}\)
PHXII03:CURRENT ELECTRICITY

356885 Two cells of internal resistances \({r_{1\,}}\,{\rm{and}}\,{r_2}\) and of same emf are connected in series across a resistor of resistance \(R\). If the terminal potential difference across the cell of internal resistance \({r_{1\,}}\) is zero, then the value of \(R\) is

1 \(R = 2({r_1} + {r_2})\)
2 \(R = {r_2} - {r_1}\)
3 \(R = {r_1} - {r_2}\)
4 \(R = 2({r_1} - {r_2})\)
PHXII03:CURRENT ELECTRICITY

356886 Ten identical cells each emf \(2 V\) and internal resistances \(1 \Omega\) are connected in series with two cells wrongly connected. A resistor of \(10 \Omega\) is connected to the combination. What is the current through the resistor?

1 \(0.6\;A\)
2 \(1.2\;A\)
3 \(1.8\;A\)
4 \(2.4\;A\)
PHXII03:CURRENT ELECTRICITY

356887 While connecting 5 cells in a battery in series, in a tape recorder, by mistake one cell is connected with reverse polarity. If the effective load resistance is \(7\,\Omega \) of load is \(24\,ohm\) and internal resistance of each cell is one ohm and emf \(2\;V\), the current delivered by the battery is

1 \(0.1\,A\)
2 \(0.5\,A\)
3 \(0.3\,A\)
4 \(0.4\,A\)
PHXII03:CURRENT ELECTRICITY

356884 Five cells each of emf \(E\) and internal resistance \(r\) send the same amount of current through an external resistance \(R\) whether the cells are connected in parallel or in series. Then the ratio \(\left( {\frac{R}{r}} \right)\)is

1 \(\frac{1}{2}\)
2 \(2\)
3 \(1\)
4 \(\frac{1}{5}\)
PHXII03:CURRENT ELECTRICITY

356885 Two cells of internal resistances \({r_{1\,}}\,{\rm{and}}\,{r_2}\) and of same emf are connected in series across a resistor of resistance \(R\). If the terminal potential difference across the cell of internal resistance \({r_{1\,}}\) is zero, then the value of \(R\) is

1 \(R = 2({r_1} + {r_2})\)
2 \(R = {r_2} - {r_1}\)
3 \(R = {r_1} - {r_2}\)
4 \(R = 2({r_1} - {r_2})\)
PHXII03:CURRENT ELECTRICITY

356886 Ten identical cells each emf \(2 V\) and internal resistances \(1 \Omega\) are connected in series with two cells wrongly connected. A resistor of \(10 \Omega\) is connected to the combination. What is the current through the resistor?

1 \(0.6\;A\)
2 \(1.2\;A\)
3 \(1.8\;A\)
4 \(2.4\;A\)
PHXII03:CURRENT ELECTRICITY

356887 While connecting 5 cells in a battery in series, in a tape recorder, by mistake one cell is connected with reverse polarity. If the effective load resistance is \(7\,\Omega \) of load is \(24\,ohm\) and internal resistance of each cell is one ohm and emf \(2\;V\), the current delivered by the battery is

1 \(0.1\,A\)
2 \(0.5\,A\)
3 \(0.3\,A\)
4 \(0.4\,A\)