When switch Sin Fig. E25.29 is open, the voltmeter V reads 3.08 V. When the switch is closed, the voltmeter reading drops to 2.97 V, and the ammeter A reads 1.65 A. Find the emf, the internal resistance of the battery, and the circuit resistance R. Assume that the two meters are ideal, so they don’t affect the circuit.

Fig. E25.29.

Short Answer

Expert verified
  1. The emf of the battery is 3.08 V.
  2. The internal resistance of the battery is0.067Ω.
  3. The circuit resistance R is 1.80Ω.

Step by step solution

01

Determination of the emf when the circuit is open.

When the switch is open, the circuit is incomplete, and therefore no current flows as shown in the figure below.

So, l = 0

The voltage drop across the circuit is

Vab=ε=3.08V

02

Determination of the emf when the circuit is closed, internal resistance and circuit resistance.

When the switch is closed, the circuit is complete and therefore current will flow as shown in figure below,

VoltagedropVab=ε-lrThevalueofVabgivenis2.97VSolveforrandsubstituteallthevalues,r=ε-Vabl=3.08V-2.97V1.65A=0.067Ω

For, circuit resistance, according to Ohm’s law,

V=lRR=Vabl=2.97V1.65A=1.80ΩThus,theinternalresistanceandthecircuitresistanceare0.067Ωand1.80Ωrespectively.

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