Respuesta :
This question involves the concepts of current, voltage, and power.
a. The ammeter should give a reading of "0.2 A".
b. The voltmeter readings should be "V₁ = 7 volts" and "V₂ = 3 volts".
c. The power delivered to resistors will be "P₁ = 1.4 W" and "P₂ = 0.6 W".
d. Energy delivered per hour will be "E₁ = 5040 J/hr" and "E₂ = 2160 J/hr".
a. Ammeter Reading
The ammeter will give the amount of total current in the circuit. This is a series circuit, so the current will be the same for all resistors. Applying Ohm's Law to the circuit gives:
V = IR
where,
- V = battery voltage = 10 v
- I = current reading = ?
- R = total resistance = R₁ + R₂ = 35 Ω + 15 Ω = 50 Ω
Therefore,
[tex]10\ v = I(50\ \Omega)\\\\I=\frac{10\ v}{50\ \Omega}\\\\[/tex]
I = 0.2 A
b. Voltmeter Readings
Voltmeter readings can be found using Ohm's law:
V₁ = IR₁ = (0.2 A)(35 Ω)
V₁ = 7 volts
V₂ = IR₂ = (0.2 A)(15 Ω)
V₂ = 3 volts
c. Power Delivered
The power delivered to each resistor can be found as follows:
P₁ = IV₁ = (0.2 A)(7 volts)
P₁ = 1.4 W
P₂ = IV₂ = (0.2 A)(3 volts)
P₂ = 0.6 W
d. Energy per hour
The energy per hour can be found as follows:
E₁ = P₁(3600 s/hr) = (1.4 W)(3600 s/hr)
E₁ = 5040 J/hr
E₂ = P₂(3600 s/hr) = (0.6 W)(3600 s/hr)
E₂ = 2160 J/hr
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