1.1 A. An electric oven with a resistance of 201Ω and a voltage of 220V drwa a current of 1.1 A.
The easiest way to solve this problem is using the Ohm's Law I = V/R.
An electric oven has R = 201Ω, and a drop of voltage V = 220v, solve using I = V/R:
I = 220V / 201Ω
I = 1.09 A ≅ 1.1 A
Answer:
Voltage across the capacitor is 30 V and rate of energy across the capacitor is 0.06 W
Explanation:
As we know that the current in the circuit at given instant of time is
i = 2.0 mA
R = 10 k ohm
now we know by ohm's law



so voltage across the capacitor + voltage across resistor = V


Now we know that

here rate of change in energy of the capacitor is given as



Answer:
I believe it is C. Their Temps.
Explanation:
Hope my answer has helped you!
Taha xain malai ..........hhdd
I agree the best choice is A............