How does changing the potential difference in a circuit affect the current and the resistance?
1 answer:
Answer:
The current will be increased and also for the resistance.
Explanation:
The analysis of a direct current circuit can give us the explanation we need. Using the ohm law, which tells us that the voltage is equal to the product of the current by the resistance we have:
The voltage is equal to the potential difference therefore we will have these expressions:
If we increase the potential differential or circuit voltage, the current will also increase and so does the resistance by increasing the voltage. If we put numerical values in the equation given before, we can confirm this fact.
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Answer:
a) I = 3.63 W / m² , b) I = 0.750 W / m²
Explanation:
The intensity of a sound wave is given by the relation
I = P / A = ½ ρ v (2π f )²
I = (½ ρ v 4π² s_{max}²) f²
a) with the initial condition let's call the intensity Io
cte = (½ ρ v 4π² s_{max}²)
I₀ = cte s² f₀²
I₀ = cte 10 6
If frequency is increase f = 2.20 10³ Hz
I = constant (2.20 10³) 2
I = cte 4.84 10⁶
let's find the relationship of the two quantities
I / Io = 4.84
I = 4.84 Io
I = 4.84 0.750
I = 3.63 W / m²
b) in this case the frequency is reduced to f = 0.250 10³ Hz and the displacement s = 4 s or
I = cte (f s)²
I = constant (0.250 10³ 4)²
I = cte 1 10⁶
the relationship
I / Io = 1
I = Io
I = 0.750 W / m²
It might make more sense putting it another way but this is basically it. you just take the minutes and divide them by 60 to convert them to hours. then simplify the ratio
