Question

When the switch S is open, the voltmeter V of the battery 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 of the battery​

Answer 1

Answer:

When switch S in the figure is open, the voltmeter V of the battery reads 3.13V . When the switch is closed, the voltmeter reading drops to 2.90V , and the ammeter A reads 1.70A. Assume that the two meters are ideal, so they do not affect the circuit.

emf= V

sorry I left out the V and the A earlier.

Answer 2

Answer:

The emf of battery is 3.08V and internal resistance is 0.067Ω.

Explanation:

Given when the switch is open, the voltmeter reading, $V = 3.08 V$

When the switch is closed, the voltmeter reading $V = 2.97 V$

Ammeter reading,  $I= 1.65 A$

Since battery is also connected in the circuit, using Ohm's law that includes the electromotive force(E) of the battery,

$V=E-Ir$

where r is the internal resistance.

When switch is opened, there will no current  flow through the circuit, and hence $I=0$

Therefore, $V=E=3.08V$

Therefore, the emf of battery is 3.08V.

In addition, to find the internal resistance of the battery, consider the case when switch is closed,

then $V=E-Ir\implies 2.97=3.08-1.65\times r$

$\implies r=\frac{3.08-2.97}{1.65} =0.067\Omega$

Therefore, internal resistance is 0.067Ω.