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Standard XII

Physics

Mixed Combination of Cells

Ten cells eac...

Question

Ten cells each of emf 1 V and internal resistance 1 $Ω$ are connected in series. In this arrangement polarity of two cells is reversed and the system is connected to an external resistance of 2 $Ω$ . Find the current in the circuit.

\frac{1}{4} A

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\frac{1}{2} A

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\frac{1}{12} A

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\frac{3}{2} A

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Solution

The correct option is C $\frac{1}{2} A$
Given : $E = 1$ V $r = 1 Ω$ $R = 2 Ω$

Total emf of the circuit $E_{e q} = 8 E - 2 E = 6 E = 6$ V

Total resistance of the circuit $R_{e q} = 10 r + R = 10 (1) + 2 = 12 Ω$

$∴$ Current in the circuit $I = \frac{E_{e q}}{R_{e q}} = \frac{6}{12} = 0.5$ A

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Ten cells each of emf 1 V and internal resistance 1Ω are connected in series. In this arrangement polarity of two cells is reversed and the system is connected to an external resistance of 2Ω . Find the current in the circuit.

The correct option is C 12AGiven : E=1 V r=1Ω R=2ΩTotal emf of the circuit Eeq=8E−2E=6E=6 VTotal resistance of the circuit Req=10r+R=10(1)+2=12Ω∴ Current in the circuit I=EeqReq=612=0.5 A

Ten cells each of emf 1 V and internal resistance 1 $Ω$ are connected in series. In this arrangement polarity of two cells is reversed and the system is connected to an external resistance of 2 $Ω$ . Find the current in the circuit.

The correct option is C $\frac{1}{2} A$
Given : $E = 1$ V $r = 1 Ω$ $R = 2 Ω$

Total emf of the circuit $E_{e q} = 8 E - 2 E = 6 E = 6$ V

Total resistance of the circuit $R_{e q} = 10 r + R = 10 (1) + 2 = 12 Ω$

$∴$ Current in the circuit $I = \frac{E_{e q}}{R_{e q}} = \frac{6}{12} = 0.5$ A