Answer:
A) two voltages are equal, B) the two voltages are different
C) voltage is the same V = 0, D) V_{battery} = V_{Lamp} > V_{wire}
Explanation:
For this exercise, it is asked to build a circuit with the plug, the lamp and the battery, the circuit must be a series circuit
Before connecting the battery voltage is Vo as they indicate that it is ideal there are no losses, the voltages of the other elements are zero.
A) Voltages battery
a) single battery Vo
b) the battery is connected to the lamp this takes something current, but the electromotive force does not change therefore the voltage is Vo; This is because the battery's energy comes from a chemical reaction of the elements inside it.
in summary the two voltages are equal
B) voltage Lamp
a) when the bulb is single, its voltage is zero V = 0
since it has no energy
b) when in the circuit V = V₀ - V_r
where V_r is the voltage across the wire, due to its resistance, the energy for these voltages is given by the battery
in this case the two voltages are different
C) the voltage across the wire
a) unconnected V = 0
b) connected V = V₀ - V_L
where V_L is the voltage across the lamp due to its resistance
we can write the wire voltage
V_r = i R
as they indicate that the wire is ideal, its resistance is zero R = 0, consequently
V_r = 0
the voltage is the same V = 0
D) circuit voltage
For this part we must write Ohm's law for this circuit
V₀ = V_L + V_r
where V₀ is the battery voltage, V_L and V_r are the voltage of the lamp and the wire, respectively.
The voltage of each element is
V_L = i R_L
V_r = i R
This is a series circuit so the current is also constant
V₀ = i (R_L + R)
therefore the voltage in the battery is the highest V₀
the voltage across the wire if it is ideal is zero V_r = 0, if it is ideal its resistance is zero
therefore the voltage on the lamp is equal to the voltage of the battery
V_{battery} = V_{Lamp} > V_{wire}