1.

    Two cells of emf E1E_1 and E2E_2 and internal resistances r1r_1 and r2r_2 respectively are connected in parallel. The equivalent emf and internal resistance of the combination will be:

    A

    Eeq=E1+E2E_{eq} = E_1 + E_2, req=r1+r2r_{eq} = r_1 + r_2

    B

    Eeq=E1r1+E2r2E_{eq} = \frac{E_1}{r_1} + \frac{E_2}{r_2}, req=r1r2r1+r2r_{eq} = \frac{r_1 r_2}{r_1 + r_2}

    C

    Eeq=E1r2+E2r1r1+r2E_{eq} = \frac{E_1 r_2 + E_2 r_1}{r_1 + r_2}, req=r1r2r1+r2r_{eq} = \frac{r_1 r_2}{r_1 + r_2}

    D

    Eeq=E1r1+E2r2E_{eq} = \frac{E_1}{r_1} + \frac{E_2}{r_2}, req=r1+r2r_{eq} = r_1 + r_2

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    2.

    A cell of e.m.f.E{\rm{e}}{\rm{.m}}{\rm{.f}}{\rm{.}}\,\,E connected with an external resistance RR, then p.d. across cell is VV. The internal resistance of cell will be

    A

    (EV)RE\frac{{\left( {E - V} \right)R}}{E}

    B

    (EV)RV\frac{{\left( {E - V} \right)R}}{V}

    C

    (VE)RV\frac{{\left( {V - E} \right)R}}{V}

    D

    (VE)RE\frac{{\left( {V - E} \right)R}}{E}

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    3.

    A battery of emfemf E{\rm{E}} produces currents I1{I_1} and I2{I_2} when connected to external resistances R1{R_1} and R2{R_2} respectively. The internal resistance of the battery is

    A

    I1R2I2R1I2I1\frac{{{I_1}\,{R_2} - {I_2}\,{R_1}}}{{{I_2} - {I_1}}}

    B

    I1R2+I2R1I1I2\frac{{{I_1}\,{R_2} + \,\,{I_2}\,{R_1}}}{{{I_1} - {I_2}}}

    C

    I1R1+I2R2I1I2\frac{{{I_1}\,{R_1} + \,\,{I_2}\,{R_2}}}{{{I_1} - {I_2}}}

    D

    I1R1I2R2I2I1\frac{{{I_1}\,{R_1} - {I_2}\,{R_2}}}{{{I_2} - {I_1}}}

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    9.

    A voltmeter having a resistance of 998ohm998\,\,ohm is connected to a cell of e.m.f. 2volt2\,\,volt and internal resistance 2ohm2\,\,ohm. The error in the measurement of e.m.f. will be

    A

    4imes101volt4\,\, imes \,\,{10^{ - 1}}\,\,volt

    B

    2imes103volt2\,\, imes \,\,{10^{ - 3}}\,\,volt

    C

    4imes103volt4\,\, imes \,\,{10^{ - 3}}\,\,volt

    D

    2imes101volt2\,\, imes \,\,{10^{ - 1}}\,\,volt

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