When switch S in the figure is open, the voltmeter Vof the battery reads 3.12 V.

Question

When switch S in the figure is open, the voltmeter Vof the battery reads 3.12 V. Whenthe switch is closed, the voltmeter reading drops to2.97 V, andthe ammeter A reads 1.65 A.Assume that the two meters are ideal, so they do not affect thecircuit a) Find the emf varepsilon. Express your answer in volts to three significant digits. b) Find the circuit resistanceR. Expressyour answer in ohms to three significant digits.

Explanation / Answer

We know , EMF is the open circuit voltaf=ge across a cell . so here when switch S is open , there is no current in the circuit, So EMF = 3.12 V Also when switch S is closed . Current in circuit = EMF / (R+r) =1.65 SO 3.12 /(r+R) = 1.65 =>R+r= 1.89 Also potential across the battery = EMF - ir =2.97 So 3.12 - 1.65*r = 2.97 r = 0.091 SO R = 1.89 - 0.091 = 1.799

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