Respuesta :
Answer:
The current in the circuit is 0.15 Ampere
Explanation:
The given parameters of the cell are;
The electromotive force (e.m.f.) of the cell, E = 1.5 V
The resistance of the cell, r = 2.5 ohm
The resistance of the ammeter = 0.5 ohm
The resistance of the resistor = 7.0 ohm
The formula for the e.m.f., E of a cell is given as follows;
e.m.f. E = I·(R + r)
Where;
I = The current in the circuit
R = The sum of the resistances in the circuit = 7.0 Ω + 0.5 Ω + 2.5 Ω = 10 Ω
Therefore, we have;
[tex]The \ current \ in \ the \ circuit, \ I = \dfrac{E}{R + r}[/tex]
Substituting the known values, gives;
[tex]I = \dfrac{1.5 \ V}{7 \ \Omega + 0.5 \ \Omega + 2.5 \ \Omega} = \dfrac{1.5 \ V}{10 \ \Omega} = 0.15 \ A[/tex]
The current in the circuit, I = 0.15 Ampere.
